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	<title>Michi's blog &#187; Algebraic geometry</title>
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		<title>Introduction to Algebraic Geometry (3 in a series)</title>
		<link>http://blog.mikael.johanssons.org/archive/2008/03/introduction-to-algebraic-geometry-3-in-a-series/</link>
		<comments>http://blog.mikael.johanssons.org/archive/2008/03/introduction-to-algebraic-geometry-3-in-a-series/#comments</comments>
		<pubDate>Tue, 04 Mar 2008 14:37:58 +0000</pubDate>
		<dc:creator>Michi</dc:creator>
				<category><![CDATA[Algebra]]></category>
		<category><![CDATA[Algebraic geometry]]></category>
		<category><![CDATA[Geometry]]></category>
		<category><![CDATA[Mathematics]]></category>

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		<description><![CDATA[I&#8217;m going to move on with the identification of geometric objects with functions from these objects down to a field soon enough, but I&#8217;d like to spend a little time nailing down the categorical language of this association. Basically, we have two functors I and V going back and forth between two categories. And the [...]]]></description>
			<content:encoded><![CDATA[<p>I&#8217;m going to move on with the identification of geometric objects with functions from these objects down to a field soon enough, but I&#8217;d like to spend a little time nailing down the categorical language of this association. Basically, we have two functors I and V going back and forth between two categories. And the essential statement of the last post is that these two functors form an equivalence of categories.</p>
<p>Now, first off in this categorical language, I want to nail down exactly what the objects are. In the category <img src='/latexrender/pictures/ad9c5aca753932fe6e45672e1941d82f.png' title='\mathcal{AV}ar_k' alt='\mathcal{AV}ar_k' align='middle' /> the objects are solution sets of systems of polynomial equations. And in the category <img src='/latexrender/pictures/b5655648e330b813df29c47685b3f653.png' title='\mathcal{RA}lg_k' alt='\mathcal{RA}lg_k' align='middle' />, the objects are finitely presented Noetherian reduced k-algebras.</p>
<p>The functor <img src='/latexrender/pictures/ad2bbacf81ace3b6fd39e64b2f038a79.png' title='V:\mathcal{RA}lg_k\to \mathcal{AV}ar_k' alt='V:\mathcal{RA}lg_k\to \mathcal{AV}ar_k' align='middle' /> acts on objects by sending an algebra R to the solution set of the polynomial equations generating the ideal in a presentation of the algebra.</p>
<p>And the functor <img src='/latexrender/pictures/5e921306cbfd443eeb1a2ef868175e24.png' title='I:\mathcal{AV}ar_k\to \mathcal{RA}lg_k' alt='I:\mathcal{AV}ar_k\to \mathcal{RA}lg_k' align='middle' /> takes a variety to the algebra of polynomial functions on the variety. This is a slight modification to the way I&#8217;ve introduced I in the previous posts &#8211; but the good news is that I(V) is the quotient of the right polynomial ring with the previously defined I(V).</p>
<h2>Morphisms on varieties</h2>
<p>In order to define a category, it&#8217;s not enough with the objects &#8211; we want morphisms as well. Since everything else is defined by polynomials, we&#8217;re going to define a morphism of varieties <img src='/latexrender/pictures/88bb990cfa90788a59705302ee7b5697.png' title='V\to W' alt='V\to W' align='middle' /> to be a map <img src='/latexrender/pictures/77abf6ed51ee3ba7e6a4964c968274ab.png' title='\mathbb A^n\to\mathbb A^m' alt='\mathbb A^n\to\mathbb A^m' align='middle' />, polynomial in each coordinate, such that the image of the restriction to the variety <img src='/latexrender/pictures/a0d9814e229150f7347974a01bd9885f.png' title='V\subseteq\mathbb A^n' alt='V\subseteq\mathbb A^n' align='middle' /> is contained in <img src='/latexrender/pictures/03cce3ca93d255d6e005d6df9043d412.png' title='W\subseteq\mathbb A^m' alt='W\subseteq\mathbb A^m' align='middle' />.</p>
<p>In other words, it is a map <img src='/latexrender/pictures/ff558b3aaadb6754a119dd20600734a6.png' title='(x_1,\dots,x_n)\mapsto(f_1(x_1,\dots,x_n),\dots,f_m(x_1,\dots,x_n))' alt='(x_1,\dots,x_n)\mapsto(f_1(x_1,\dots,x_n),\dots,f_m(x_1,\dots,x_n))' align='middle' /> defined by polynomials <img src='/latexrender/pictures/312d9bcb125a3799fad4b84f6991b12e.png' title='f_1,\dots,f_m' alt='f_1,\dots,f_m' align='middle' />.</p>
<p>An isomorphism of varieties is exactly what we expect it to be &#8211; it is a morphism that has an inverse.</p>
<p>One specific kind of highly interesting isomorphisms are the linear automorphisms of <img src='/latexrender/pictures/e75c212414a6a6750f34b305153b0648.png' title='\mathbb A^n' alt='\mathbb A^n' align='middle' />. These are given, essentially, by invertible change-of-coordinate matrices in the way we are used to from linear algebra. </p>
<h3>Examples</h3>
<p>Recall from the earlier posts the twisted cubic curve <img src='/latexrender/pictures/a821c4a9f5431350d01ff39b028e2a3d.png' title='V(x-y^2,x-z^3)' alt='V(x-y^2,x-z^3)' align='middle' />. Points on it have the form <img src='/latexrender/pictures/19f66dbddba8ea791961821590ae1d5e.png' title='(s,s^2,s^3)' alt='(s,s^2,s^3)' align='middle' /> &#8211; and this, incidentially, gives us precisely a map <img src='/latexrender/pictures/f406f0a4b8c4fa5e07359e57d824aab5.png' title='\mathbb A^1\to\mathbb A^3' alt='\mathbb A^1\to\mathbb A^3' align='middle' /> displaying an isomorphism between the twisted cubic curve and the affine line. The inverse is given by <img src='/latexrender/pictures/8e6e7f06e5038c6ccb179b83f4c76c68.png' title='(x,y,z)\mapsto x' alt='(x,y,z)\mapsto x' align='middle' />. </p>
<p>Consider the parabola <img src='/latexrender/pictures/49234b47786c5e019eec30ef3f77019a.png' title='V(x-y^2)' alt='V(x-y^2)' align='middle' />. This is also isomorphic to the affine line, over the maps <img src='/latexrender/pictures/d0b89cda9761a4603b976aa6d5a04b39.png' title='t\mapsto(t,t^2)' alt='t\mapsto(t,t^2)' align='middle' /> and <img src='/latexrender/pictures/32edafbf1ff3665f4b9cd387f1e81d8c.png' title='(x,y)\mapsto x' alt='(x,y)\mapsto x' align='middle' />.</p>
<p>On the other hand, the affine line is not isomorphic to the nodal curve <img src='/latexrender/pictures/e313f494986a12def52f0e61484c018f.png' title='V(y^2-x^2-x^3)' alt='V(y^2-x^2-x^3)' align='middle' />. The easiest way to show this is to go over smoothness of curves and singular points &#8211; which I hope to deal with later at some point. Essentially, smoothness is an invariant of varieties under isomorphisms, and since the point (0,0) is singular on the nodal curve, and the affine line has no singular points, the two varieties can not possibly be isomorphic.</p>
<p>Note that images of varieties need not be affine algebraic varieties &#8211; they will, however, always be <i>quasi-projective</i> varieties. We&#8217;ll see if I get into this later on.</p>
<h2>Morphisms of algebras and functoriality of V and I</h2>
<p>We really do already know what morphisms look like in <img src='/latexrender/pictures/b5655648e330b813df29c47685b3f653.png' title='\mathcal{RA}lg_k' alt='\mathcal{RA}lg_k' align='middle' />. This category is the full subcategory of the category of k-algebras &#8211; by which we mean that it picks out objects among k-algebras, and have all k-algebra maps between objects as morphisms.</p>
<p>The really awesome bit happens when we start considering the morphisms we&#8217;ve defined. Given a morphism <img src='/latexrender/pictures/47b3a34fa2ec6da6d58a234edbfd0d7b.png' title='F:V\to W' alt='F:V\to W' align='middle' />, we define the pullback <img src='/latexrender/pictures/58d9d4ce1976a966882e7c82d82b3b16.png' title='F^\#:k[W]\to k[V]' alt='F^\#:k[W]\to k[V]' align='middle' /> by <img src='/latexrender/pictures/c21edb1fa4fc99d6f96144f3abbad346.png' title='F^\#(f)=f\circ F' alt='F^\#(f)=f\circ F' align='middle' />. This takes a map <img src='/latexrender/pictures/0a5561ce152d5d314411bb9f4df4718d.png' title='f:W\to k' alt='f:W\to k' align='middle' /> and makes a map <img src='/latexrender/pictures/7d5999df6f5822fbdbd7e461e5614664.png' title='F^\#(f):V\to k' alt='F^\#(f):V\to k' align='middle' />. Since this is a composition of polynomials, it is also a polynomial function. If <img src='/latexrender/pictures/99089dce014558ee490160cba954f89e.png' title='f\in I(V)' alt='f\in I(V)' align='middle' />, then <img src='/latexrender/pictures/6ba86ecf71bf4f9bc3b3507743e71970.png' title='F^\#(f)(p)=f(F(p))' alt='F^\#(f)(p)=f(F(p))' align='middle' />, and since <img src='/latexrender/pictures/c9ecad04fff1e0a26cd5ab52b4e4264a.png' title='F(p)\in V' alt='F(p)\in V' align='middle' />, it follows that <img src='/latexrender/pictures/65cef750edb74dcc5a332a39cf4bf100.png' title='f(F(p))=0' alt='f(F(p))=0' align='middle' />, and thus <img src='/latexrender/pictures/b90b16491539207246328339b7815826.png' title='F^\#(f)\in I(W)' alt='F^\#(f)\in I(W)' align='middle' />.</p>
<p>In the other direction, suppose that R and S are reduced finitely generated k-algebras. Then <img src='/latexrender/pictures/a9e1d837f0532588473cb1ab5c2018c2.png' title='R=k[x_1,\dots,x_r]/I' alt='R=k[x_1,\dots,x_r]/I' align='middle' />, and <img src='/latexrender/pictures/10466ee4c934178961c960b6eb8d00df.png' title='S=k[y_1,\dots,y_s]/J' alt='S=k[y_1,\dots,y_s]/J' align='middle' />. We fix a homomorphism <img src='/latexrender/pictures/9a137544a74c2d34def9f6ada25d0fa0.png' title='\sigma:R\to S' alt='\sigma:R\to S' align='middle' />, and we wish to construct a variety morphism <img src='/latexrender/pictures/800618943025315f869e4e1f09471012.png' title='F' alt='F' align='middle' /> such that <img src='/latexrender/pictures/355d967e735aa975780820018f9fa12d.png' title='F^\#=\sigma' alt='F^\#=\sigma' align='middle' />.</p>
<p>Let <img src='/latexrender/pictures/824ff39cf03c28e185367a8974a30aeb.png' title='F_i\in k[y_1,\dots,y_s]' alt='F_i\in k[y_1,\dots,y_s]' align='middle' /> be a representative of <img src='/latexrender/pictures/f35d2abca0690e7b10a5f2bb687554e2.png' title='\sigma(x_i)' alt='\sigma(x_i)' align='middle' />, and define <img src='/latexrender/pictures/04490679b59f37969f27d96ed4d006e6.png' title='F:\mathbb A^s\to \mathbb A^r' alt='F:\mathbb A^s\to \mathbb A^r' align='middle' /> by <img src='/latexrender/pictures/0052ae45694c9a8049b97f0611d1bc89.png' title='a\mapsto(F_1(a),\dots,F_r(a))' alt='a\mapsto(F_1(a),\dots,F_r(a))' align='middle' />. We need to verify that F maps V to W. This follows if we can only show that for every <img src='/latexrender/pictures/74b772ed1f57705570a5d329409a237f.png' title='a\in V' alt='a\in V' align='middle' />, all polynomials in I vanish on F(a). Let <img src='/latexrender/pictures/dd8b6e371f0037673846188fda7a78ec.png' title='g\in I' alt='g\in I' align='middle' />. Then<br />
<img src='/latexrender/pictures/3085e2eef4109a4f710816b4afda4c87.png' title='g(F(a))=g(F_1(a),\dots,F_r(a))=g(\sigma(x_1)(a),\dots,\sigma(x_r)(a))=\sigma(g)(a)' alt='g(F(a))=g(F_1(a),\dots,F_r(a))=g(\sigma(x_1)(a),\dots,\sigma(x_r)(a))=\sigma(g)(a)' align='middle' /><br />
and since <img src='/latexrender/pictures/dd8b6e371f0037673846188fda7a78ec.png' title='g\in I' alt='g\in I' align='middle' />, it represents the zero class of <img src='/latexrender/pictures/94480f1b196cbb24f2c8bacf58bccb5f.png' title='k[x_1,\dots,x_r]/I' alt='k[x_1,\dots,x_r]/I' align='middle' />, so <img src='/latexrender/pictures/7e0fb729f38d2e86c8580d5abddfe08f.png' title='\sigma(g)=0' alt='\sigma(g)=0' align='middle' /> and hence <img src='/latexrender/pictures/6ba5f00aca6154941916eea5de601783.png' title='\sigma(g)\in J' alt='\sigma(g)\in J' align='middle' />. But J is the ideal of all functions vanishing on V. Hence <img src='/latexrender/pictures/32c6c0c4bca4004c0b77bf54a50b8547.png' title='F(a)\in W' alt='F(a)\in W' align='middle' />, and the proof is complete.</p>
<p>In essence, what this proves to us is that the operations V and I form a contravariant equivalence of categories between <img src='/latexrender/pictures/1b8e7d068548e01cd73072b4dbffdb3a.png' title='V:\mathcal{RA}lg_k' alt='V:\mathcal{RA}lg_k' align='middle' /> and <img src='/latexrender/pictures/ad9c5aca753932fe6e45672e1941d82f.png' title='\mathcal{AV}ar_k' alt='\mathcal{AV}ar_k' align='middle' />.</p>
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		<title>Introduction to Algebraic Geometry (2 in a series)</title>
		<link>http://blog.mikael.johanssons.org/archive/2008/02/introduction-to-algebraic-geometry-2-in-a-series/</link>
		<comments>http://blog.mikael.johanssons.org/archive/2008/02/introduction-to-algebraic-geometry-2-in-a-series/#comments</comments>
		<pubDate>Thu, 21 Feb 2008 21:43:46 +0000</pubDate>
		<dc:creator>Michi</dc:creator>
				<category><![CDATA[Algebra]]></category>
		<category><![CDATA[Algebraic geometry]]></category>
		<category><![CDATA[Geometry]]></category>
		<category><![CDATA[Mathematics]]></category>

		<guid isPermaLink="false">http://blog.mikael.johanssons.org/archive/2008/02/introduction-to-algebraic-geometry-2-in-a-series/</guid>
		<description><![CDATA[I want to lead this sequence to the point where I am having trouble understanding algebraic geometry. Hence, I won&#8217;t take the usual course such an introduction would take, but rather set the stage reasonably quickly to make the transit to the more abstract themes clear. But that&#8217;s all a few posts away. For now, [...]]]></description>
			<content:encoded><![CDATA[<p>I want to lead this sequence to the point where I am having trouble understanding algebraic geometry. Hence, I won&#8217;t take the usual course such an introduction would take, but rather set the stage  reasonably quickly to make the transit to the more abstract themes clear.</p>
<p>But that&#8217;s all a few posts away. For now, recall that we recognized already that any variety is defined by an ideal, and that intersections and unions of varieties are given by sums and intersections or products of ideals.</p>
<p>This is the first page of what is known as the Algebra-Geometry dictionary. The dictionary is made complete by a pair of reasonably famous theorems. I won&#8217;t bother proving them &#8211; the proofs are a good chunk of any decent commutative algebra course &#8211; but I&#8217;ll quote the theorems and discuss why they matter.</p>
<p>We call a ring Noetherian if all ideals are finitely generated. If a ring R is Noetherian, then quotients are Noetherian.</p>
<p><b>Hilbert&#8217;s Basis Theorem:</b> If R is Noetherian, then so is R[x].</p>
<p>We define the radical <img src='/latexrender/pictures/4bd65db8055fe5cdb1ee85f1073c8661.png' title='\sqrt I' alt='\sqrt I' align='middle' /> of an ideal I to be the ideal consisting of all elements a such that some power of a is actually in I. We call an ideal I <i>radical</i> if <img src='/latexrender/pictures/7968b9f14c3a27f504c104cd11f19a60.png' title='I=\sqrt I' alt='I=\sqrt I' align='middle' />. This concept is relevant for our considerations since if for a point p the function f^n(p) vanishes, then f(p) also vanishes. Thus, the set of points such that f^n vanishes is the same set as the set of points where f vanishes. The relevancy of this is captured in:</p>
<p><b>Hilbert&#8217;s Nullstellensatz:</b> Let k be an algebraically closed field. For any ideal I in <img src='/latexrender/pictures/1de8377bc00a12d8d4e4641774350142.png' title='k[x_1,\dots,x_n]' alt='k[x_1,\dots,x_n]' align='middle' />, there is an equality of ideals<br />
<img src='/latexrender/pictures/a900c9759a5cbcc1f3e700dd110e7e5c.png' title='I(V(I))=\sqrt I' alt='I(V(I))=\sqrt I' align='middle' /></p>
<p>Note, for the statement of this theorem that we write V(I) for the variety defined by simultaneous vanishing of all elements in I, and we write I(V) for the ideal of all polynomials in <img src='/latexrender/pictures/1de8377bc00a12d8d4e4641774350142.png' title='k[x_1,\dots,x_n]' alt='k[x_1,\dots,x_n]' align='middle' /> that vanish on all of I(V). </p>
<p>So &#8211; and here is the beautiful part &#8211; affine algebraic varieties correspond bijectively to radical ideals in polynomial rings. For every ideal, there is a variety and for every variety, there is an ideal. But we can push this further.</p>
<h2>Coordinate rings</h2>
<p>Let&#8217;s consider polynomial functions from <img src='/latexrender/pictures/e75c212414a6a6750f34b305153b0648.png' title='\mathbb A^n' alt='\mathbb A^n' align='middle' /> to k. These are precisely the polynomials in <img src='/latexrender/pictures/1de8377bc00a12d8d4e4641774350142.png' title='k[x_1,\dots,x_n]' alt='k[x_1,\dots,x_n]' align='middle' />.  Given a variety V, we can take a polynomial <img src='/latexrender/pictures/7346c23f9188b822dc31297afce98edf.png' title='f\in k[x_1,\dots,x_n]' alt='f\in k[x_1,\dots,x_n]' align='middle' /> and restrict it to a function <img src='/latexrender/pictures/3a8276462b14ee9342582be8695e0872.png' title='f|_V:V\to k' alt='f|_V:V\to k' align='middle' />.</p>
<p>Two different polynomials give the same restricted function precisely when their difference vanishes on all of V. So polynomial functions on V are precisely the equivalence classes in the quotient ring <img src='/latexrender/pictures/54d7d4e6c0660e03259875db941611fd.png' title='k[x_1,\dots,x_n]/I(V)' alt='k[x_1,\dots,x_n]/I(V)' align='middle' />. We call the resulting ring the <i>coordinate ring</i> and denote it by k[V].</p>
<p>Conversely, if R is a Noetherian k-algebra such that there are no nilpotent elements in R, then R is a quotient of some polynomial ring with some radical ideal. Hence it is the coordinate ring of some variety in some affine space somewhere. We call a ring lacking nilpotents <i>reduced</i>. </p>
<p>We get, out of all this, a bijective correspondence<br />
{ Noetherian reduced k-algebras } <img src='/latexrender/pictures/19401793ec61890d228748bce42da457.png' title='\leftrightarrow' alt='\leftrightarrow' align='middle' /> { Affine algebraic varieties }</p>
<p>The really beautiful part will come in my next post. We can introduce homomorphisms of varieties in a reasonably natural way so that this bijective correspondence ends up being <i>functorial</i> &#8211; i.e. any homomorphisms on one side gives rise to a corresponding homomorphism on the other side. Thus, the categories of Noetherian reduced k-algebras and of affine algebraic varieties are equivalent.</p>
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		<title>Introduction to Algebraic Geometry (1 in a series)</title>
		<link>http://blog.mikael.johanssons.org/archive/2008/02/introduction-to-algebraic-geometry-1-in-a-series/</link>
		<comments>http://blog.mikael.johanssons.org/archive/2008/02/introduction-to-algebraic-geometry-1-in-a-series/#comments</comments>
		<pubDate>Thu, 21 Feb 2008 11:33:00 +0000</pubDate>
		<dc:creator>Michi</dc:creator>
				<category><![CDATA[Algebra]]></category>
		<category><![CDATA[Algebraic geometry]]></category>
		<category><![CDATA[English]]></category>
		<category><![CDATA[Geometry]]></category>
		<category><![CDATA[Mathematics]]></category>

		<guid isPermaLink="false">http://blog.mikael.johanssons.org/archive/2008/02/introduction-to-algebraic-geometry-1-in-a-series/</guid>
		<description><![CDATA[I&#8217;m growing embarrassed by my lack of understanding for the sheaf-theoretic approaches to algebraic (and differential) geometry. I&#8217;ve tried to deal with it several times before, and I&#8217;m currently reading up on Algebraic Geometry again to fill the void that the finished thesis, soon arriving travels and non-existent job application responses produce. So, why not [...]]]></description>
			<content:encoded><![CDATA[<p>I&#8217;m growing embarrassed by my lack of understanding for the sheaf-theoretic approaches to algebraic (and differential) geometry. I&#8217;ve tried to deal with it several times before, and I&#8217;m currently reading up on Algebraic Geometry again to fill the void that the finished thesis, soon arriving travels and non-existent job application responses produce.</p>
<p>So, why not learn by teaching? It&#8217;s an approach that has been pretty darn good in the past. So I thought I&#8217;d write a sequence of posts on algebraic geometry, introducing what it&#8217;s supposed to be about and how the main viewpoints develop more or less naturally from the approaches taken.</p>
<h2>Varieties</h2>
<p>The basic objective of algebraic geometry is to study solution sets to systems of polynomial equations. That is, we take some set <img src='/latexrender/pictures/1390c486ce69cc6b0865fafef40a25b1.png' title='f_1,\dots,f_r' alt='f_1,\dots,f_r' align='middle' /> of polynomials in some polynomial ring <img src='/latexrender/pictures/1de8377bc00a12d8d4e4641774350142.png' title='k[x_1,\dots,x_n]' alt='k[x_1,\dots,x_n]' align='middle' /> over some field <img src='/latexrender/pictures/8ce4b16b22b58894aa86c421e8759df3.png' title='k' alt='k' align='middle' />. And we write <img src='/latexrender/pictures/421328c5ab67bd781629a1cb95f6ae70.png' title='V(f_1,\dots,f_r)' alt='V(f_1,\dots,f_r)' align='middle' /> for the set of all simultaneous roots to all these polynomials:<br />
<img src='/latexrender/pictures/73b0e367b655a2fafacf4f0b97de31e1.png' title='V(f_1,\dots,f_r)=\{p\in k^n:f_1(p)=0, \dots, f_r(p)=0\}' alt='V(f_1,\dots,f_r)=\{p\in k^n:f_1(p)=0, \dots, f_r(p)=0\}' align='middle' /></p>
<p>If we write our polynomials as coming from the ring <img src='/latexrender/pictures/1de8377bc00a12d8d4e4641774350142.png' title='k[x_1,\dots,x_n]' alt='k[x_1,\dots,x_n]' align='middle' />, then the corresponding solution points will be points in the vector space <img src='/latexrender/pictures/aafabbf195eb3424ca815b9afe5caff1.png' title='k^n' alt='k^n' align='middle' />. In order to emphasize that we do not care for the vector space structure of this space, we shall denote it with <img src='/latexrender/pictures/e75c212414a6a6750f34b305153b0648.png' title='\mathbb A^n' alt='\mathbb A^n' align='middle' />, or if we want to emphasize the field, with <img src='/latexrender/pictures/b5f865c4e2e101f41565160abe202464.png' title='\mathbb A^n_k' alt='\mathbb A^n_k' align='middle' />. </p>
<p>The first observation at this point is that if we take the polynomial <img src='/latexrender/pictures/2cd3b198cc2d1805468e336d663e92cb.png' title='x^2+1' alt='x^2+1' align='middle' />, then the solution set over <img src='/latexrender/pictures/07e5a4a56a57f5c874ebf79bb67a0b18.png' title='\mathbb R' alt='\mathbb R' align='middle' /> is empty, while the solution set over <img src='/latexrender/pictures/b36eac3d201a1a25dda928990a79d751.png' title='\mathbb C' alt='\mathbb C' align='middle' /> is not. So, in order to set all solution sets on an equal footing &#8211; and also to make the later occurring correspondences work out &#8211; we shall require <img src='/latexrender/pictures/8ce4b16b22b58894aa86c421e8759df3.png' title='k' alt='k' align='middle' /> to be an algebraically closed field. In other words, we can always find a root to any polynomial.</p>
<p>We call the solution sets <i>varieties</i> (or &#8211; in order to distinguish from everything else we might encounter, we shall call them <i>affine algebraic varieties</i>).</p>
<p>So, the study of solutions to systems of polynomial equations is the study of varieties. And hence geometry. This neatly expands on the classical linear algebra viewpoint &#8211; where we study systems of linear equations as intersections of planes. It turns out that the main computational approach &#8211; Gröbner bases &#8211; actually specializes to the Gauß algorithm on matrices if we specialize to linear systems of equations.</p>
<h3>Examples</h3>
<p>The unit circle is an affine variety, given by the vanishing of <img src='/latexrender/pictures/3977406a9c38913879d488d5e68c6520.png' title='x^2+y^2-1' alt='x^2+y^2-1' align='middle' />. The hyperbola <img src='/latexrender/pictures/6a2d024426c0c93cbf6641b8380e2dd6.png' title='xy-1' alt='xy-1' align='middle' /> and the parabola <img src='/latexrender/pictures/30cdb631bd165a5b08ec12d1b0d4ddf1.png' title='y-x^2' alt='y-x^2' align='middle' /> are also affine varieties.</p>
<p>Another very commonly used example is the twisted cubic curve &#8211; given by the simultaneous vanishing of the parabolic sheet <img src='/latexrender/pictures/30cdb631bd165a5b08ec12d1b0d4ddf1.png' title='y-x^2' alt='y-x^2' align='middle' /> and the cubic curve sheet <img src='/latexrender/pictures/2ad4639fdd5531832be18110513c7f22.png' title='z-x^3' alt='z-x^3' align='middle' />.</p>
<h2>Changing the defining equations</h2>
<p>The points p at which both f and g vanish are the same points as those where f and f+g vanish. Indeed, if f(p)=0 and g(p)=0, then (f+g)(p)=0 as well, so the points in V(f,g) are all contained in V(f,f+g). On the other hand, suppose that f(p)=0 and (f+g)(p)=0. Then 0=(f+g)(p)=f(p)+g(p)=0+g(p). So g(p)=0 as well. Hence, all points in V(f,f+g) are contained in V(f,g). So the two varieties are equal.</p>
<p>Suppose that <img src='/latexrender/pictures/ce971a1993da574a7c38d708582ca8ea.png' title='p\in V(f,\dots)' alt='p\in V(f,\dots)' align='middle' />. Then for any h=fg we&#8217;ll see that h(p)=f(p)g(p)=0g(p)=0. Hence, if f vanishes at p, then every polynomial that f divides will also vanish at p.</p>
<p>Now, a set of polynomials in <img src='/latexrender/pictures/1de8377bc00a12d8d4e4641774350142.png' title='k[x_1,\dots,x_n]' alt='k[x_1,\dots,x_n]' align='middle' /> that is closed under addition and multiplication by elements from <img src='/latexrender/pictures/1de8377bc00a12d8d4e4641774350142.png' title='k[x_1,\dots,x_n]' alt='k[x_1,\dots,x_n]' align='middle' /> is an <i>ideal</i> of the polynomial ring. Thus, these considerations convince us that the set of all polynomials in <img src='/latexrender/pictures/1de8377bc00a12d8d4e4641774350142.png' title='k[x_1,\dots,x_n]' alt='k[x_1,\dots,x_n]' align='middle' /> that all vanish on all points in <img src='/latexrender/pictures/421328c5ab67bd781629a1cb95f6ae70.png' title='V(f_1,\dots,f_r)' alt='V(f_1,\dots,f_r)' align='middle' /> is in fact an ideal. </p>
<p>So we can define the variety from an ideal &#8211; given <img src='/latexrender/pictures/5bddc0766ff957acc660c8d5a63c6053.png' title='I\in k[x_1,\dots,x_n]' alt='I\in k[x_1,\dots,x_n]' align='middle' />, we write V(I) for the set of all points <img src='/latexrender/pictures/ed19d38fb0dbbf1787afbbf228a9133d.png' title='p\in\mathbb A^n' alt='p\in\mathbb A^n' align='middle' /> such that f(p)=0 for all <img src='/latexrender/pictures/1b4e910cd8af4b5c4899976bf945ec4b.png' title='f\in I' alt='f\in I' align='middle' />.</p>
<p>The sum of two ideals I and J are the set of all elements on the form f+g where f is from I and g is from J. If I is generated by <img src='/latexrender/pictures/1390c486ce69cc6b0865fafef40a25b1.png' title='f_1,\dots,f_r' alt='f_1,\dots,f_r' align='middle' /> and J is generated by <img src='/latexrender/pictures/f73102dc281850625bd26dd4528ae267.png' title='g_1,\dots,g_s' alt='g_1,\dots,g_s' align='middle' />, then I+J has generators <img src='/latexrender/pictures/2af066e9b2d8c917659b240f409c9260.png' title='f_1,\dots,f_r,g_1,\dots,g_s' alt='f_1,\dots,f_r,g_1,\dots,g_s' align='middle' />. And the set of points where all the polynomials in I+J vanish are precisely the points where all polynomials in I vanish <b>and</b> all polynomials in J vanish. Hence <img src='/latexrender/pictures/e0db24317effc56d994c5562d0cb448c.png' title='V(I+J)=V(I)\cap V(J)' alt='V(I+J)=V(I)\cap V(J)' align='middle' />.</p>
<p>The product IJ of two ideals I and J is <i>generated by</i> all products of one element from I and one from J. Hence, an element of IJ is on the form <img src='/latexrender/pictures/30ad3a86de244ae7f4b526c1f87e74a0.png' title='f_1g_1+\dots+f_tg_t' alt='f_1g_1+\dots+f_tg_t' align='middle' /> with all the <img src='/latexrender/pictures/59bdf0ba696e13164c5a926386f23cb0.png' title='f_i' alt='f_i' align='middle' /> coming from I and all the <img src='/latexrender/pictures/8d2d80ef00296b9ed491d989b30cebd1.png' title='g_j' alt='g_j' align='middle' /> coming from J. Suppose p is in V(IJ). Then for all f in I and all g in J, f(p)g(p)=0. If for all f in I, f(p)=0, then p is in V(I). Otherwise we can find some <img src='/latexrender/pictures/78155d2ec4b025a5ac903d6f8611756c.png' title='f_0' alt='f_0' align='middle' /> such that <img src='/latexrender/pictures/ab97a49df554c52b299d52592994a023.png' title='f_0(p)=0' alt='f_0(p)=0' align='middle' />. But then since <img src='/latexrender/pictures/29f46b78756dc0ab080236ed980a5f50.png' title='f_0(p)g(p)=0' alt='f_0(p)g(p)=0' align='middle' /> for all g in J, we must have <img src='/latexrender/pictures/a38bc5cfe1943b7eac1b3730060482cd.png' title='g(p)=0' alt='g(p)=0' align='middle' /> for all g in J. Hence p is in V(J).<br />
Thus <img src='/latexrender/pictures/b8fe0d8bd1dac25031abbc996c901662.png' title='V(IJ)\subseteq V(I)\cup V(J)' alt='V(IJ)\subseteq V(I)\cup V(J)' align='middle' />.<br />
On the other hand, if p is in <img src='/latexrender/pictures/c62bf10d4226e06e1609d3e55795b7e6.png' title='V(I)\cup V(J)' alt='V(I)\cup V(J)' align='middle' /> then specifically either p is in V(I) or it is in V(J). Either way, for all f in I and all g in J, f(p)g(p)=0 since one of the factors certainly vanishes. Hence, p is in V(IJ).</p>
<p>We also do have that <img src='/latexrender/pictures/e0c59baf5837f5962f077d0d2f7c000f.png' title='V(I\cap J)=V(I)\cup V(J)' alt='V(I\cap J)=V(I)\cup V(J)' align='middle' />. Which of the ideal operations works best depends a bit on what we would like to <i>do</i> with it.</p>
<h3>Examples</h3>
<p>We can use the above discussion to actually <i>compute</i> things geometrically. The intersection of the circle <img src='/latexrender/pictures/bc33f6de26676290a237e01b8a5a3f41.png' title='V(x^2+y^2-1)' alt='V(x^2+y^2-1)' align='middle' /> with a line <img src='/latexrender/pictures/00b84bf2a5fd7f158877522fe56f9b5a.png' title='V(ax+by+c)' alt='V(ax+by+c)' align='middle' /> is given by the sum of the ideals &#8211; so we need to consider the ideal <img src='/latexrender/pictures/b8b1a683c0c431b7e99228048d66ce9f.png' title='(x^2+y^2-1,ax+by+c)' alt='(x^2+y^2-1,ax+by+c)' align='middle' /> in <img src='/latexrender/pictures/464f66865a3de21f2f2a6a2b9a594865.png' title='k[x,y]' alt='k[x,y]' align='middle' />. We could try to bash this out by just wrangling the corresponding equations, or we could systematize the wrangling. Systematizing it leads, basically, to the theory of Gröbner bases &#8211; and this is something I tend to avoid doing by hand if I can. Not doing it by hand looks like this:</p>
<pre>
$ Singular
                     SINGULAR                             /
 A Computer Algebra System for Polynomial Computations   /   version 3-0-2
                                                       0<
     by: G.-M. Greuel, G. Pfister, H. Schoenemann        \   July 2006
FB Mathematik der Universitaet, D-67653 Kaiserslautern    \
> ring R=0,(x,y),dp;
> ideal I=x^2+y^2-1,2*x-y;
> std(I);
_[1]=2x-y
_[2]=5y2-4
</pre>
<p>So one good set of generators for the ideal <img src='/latexrender/pictures/17f0e736fb07f21e69030ae7588a2cc9.png' title='(x^2+y^2-1,2x-y)' alt='(x^2+y^2-1,2x-y)' align='middle' /> would be <img src='/latexrender/pictures/baed4e0a9225276151aa20d11c978d64.png' title='(2x-y,5y^2-4)' alt='(2x-y,5y^2-4)' align='middle' />. This, in turn corresponds to the conditions <img src='/latexrender/pictures/db51cb7add5c262eb3c95f8a9ed722f9.png' title='y=\pm 2/\sqrt 5' alt='y=\pm 2/\sqrt 5' align='middle' /> and <img src='/latexrender/pictures/989211a1ba52351d7888cc4a803604c6.png' title='x=y/2' alt='x=y/2' align='middle' />. So we get two points &#8211; just as expected.</p>
<p>The equation wrangling, on the other hand, begins with recognizing that if both a and b are 0, then V(ax+by+c) is either empty or the entire plane. Both are rather boring, and easy to handle. So one of a and b is non-zero. Let&#8217;s say that a isn&#8217;t 0. Then we can rewrite ax+by+c=0 to <img src='/latexrender/pictures/76eee6741d3651ff483e15e86c455bc3.png' title='x=-\frac ba y-\frac ca' alt='x=-\frac ba y-\frac ca' align='middle' />. This can be inserted into the quadratic equation <img src='/latexrender/pictures/0d34c87c64f32752b59c9fc56b15ad09.png' title='x^2+y^2-1=0' alt='x^2+y^2-1=0' align='middle' /> to form the new equation <img src='/latexrender/pictures/41856b97b5f02a1fae412c2b07915ab5.png' title='(-\frac ba y-\frac ca)^2+y^2=1' alt='(-\frac ba y-\frac ca)^2+y^2=1' align='middle' />, which in turn expands to<br />
<img src='/latexrender/pictures/d7c6f23b7f88c3f0af3b50b9b16effa3.png' title='(1+\frac{b^2}{a^2})y^2+2\frac{b+c}ay+\frac{c^2}{a^2}=1' alt='(1+\frac{b^2}{a^2})y^2+2\frac{b+c}ay+\frac{c^2}{a^2}=1' align='middle' /><br />
which, in turn, we can easily solve using normal solution techniques for quadratic equations. To get rid of fractions, we&#8217;ll multiply the entire equation by <img src='/latexrender/pictures/ebc3d7bedc1f11e08895c3124001cbb5.png' title='a^2' alt='a^2' align='middle' /> yielding the new equation<br />
<img src='/latexrender/pictures/0241be1468ea35c3be47c52c65dbece9.png' title='(a^2+b^2)y^2+2a(b+c)y+c^2-a^2=0' alt='(a^2+b^2)y^2+2a(b+c)y+c^2-a^2=0' align='middle' /><br />
Now, the discriminant<br />
<img src='/latexrender/pictures/5d244e36043799199bc39a52bb16820e.png' title='\Delta=(2a(b+c))^2-4(b^2+a^2)(c^2-a^2)' alt='\Delta=(2a(b+c))^2-4(b^2+a^2)(c^2-a^2)' align='middle' /><br />
carries information about the nature of solutions. If <img src='/latexrender/pictures/b8151f6178433af03a399b1971d6448c.png' title='\Delta=0' alt='\Delta=0' align='middle' />, then the line V(ax+by+c) is a tangent to the circle, if <img src='/latexrender/pictures/da44aac94b82135888c080a06999d25a.png' title='\Delta&lt;0' alt='\Delta&lt;0' align='middle' />, then the intersections are all complex and if <img src='/latexrender/pictures/3ea7fefa1f6fc0d907e126d1ee5acc34.png' title='\Delta&gt;0' alt='\Delta&gt;0' align='middle' />, then the intersections are real. In both the non-tangent cases, the circle intersects the line exactly twice. </p>
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		<title>Algebraic surface toys!</title>
		<link>http://blog.mikael.johanssons.org/archive/2008/01/algebraic-surface-toys/</link>
		<comments>http://blog.mikael.johanssons.org/archive/2008/01/algebraic-surface-toys/#comments</comments>
		<pubDate>Fri, 25 Jan 2008 16:58:27 +0000</pubDate>
		<dc:creator>Michi</dc:creator>
				<category><![CDATA[Algebra]]></category>
		<category><![CDATA[Algebraic geometry]]></category>
		<category><![CDATA[Geometry]]></category>
		<category><![CDATA[Jahr der Mathematik]]></category>

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		<description><![CDATA[At the start of the German Year of Mathematics, the Oberwolfach research institute has released an exhibition and the software they used to produce it. The software, surfer, is a really nice GUI that sits on top of surf and lets you rotate and zoom your algebraic surfaces as well as pick colours very comfortably. [...]]]></description>
			<content:encoded><![CDATA[<p>At the start of the <a href=http://www.jahr-der-mathematik.de/>German Year of Mathematics</a>, the <a href=http://www.mfo.de>Oberwolfach research institute</a> has released an exhibition and the software they used to produce it. The software, <a href=http://imaginary2008.de/surfer.php>surfer</a>, is a really nice GUI that sits on top of <a href=http://surf.sourceforge.net>surf</a> and lets you rotate and zoom your algebraic surfaces as well as pick colours very comfortably.</p>
<p>They have a whole bunch of Really Pretty Images at <a href=http://imaginary2008.de>the exhibition website</a>, and I warmly recommend a visit. If you can get hold of the exhibition, they also have produced real models &#8211; with a 3d-printer &#8211; of some of the snazzier surfaces, so that one could have a REALLY close encounter with them.</p>
<p>But also, I&#8217;d really like to show you some of my own minor experiments with the program.</p>
<p><img src="http://mikael.johanssons.org/surfer/tubamirum.png" alt="Tuba Mirum - the innards of a Klein Bottle" width="75%" /><br />
This is the interior of a Klein Bottle, using the &#8220;standard&#8221; realization as an algebraic surface given by Mathworld. In other words, I&#8217;m using<br />
(x^2+y^2+z^2+2*y-1)*((x^2+y^2+z^2-2*y-1)^2-8*z^2)+16*x*z*(x^2+y^2+z^2-2*y-1)=0<br />
for the defining equation. It kinda looks a bit like a Sousaphone in my opinion.</p>
<p><img src="http://mikael.johanssons.org/surfer/roman.png" alt="Roman's surface - immersion of the real projective plane." width="75%" /><br />
Roman&#8217;s surface, an immersion of the real projective plane into 3-dimensional euclidean space. It is given by the equation<br />
(x^2+y^2+z^2-9)^2-((z-3)^2-2*x^2)*((z+3)^2-2*y^2)=0<br />
and is one of the Steiner projections of the Veronese surface, embedding the real projective plane into projective 5-dimensional space by the homogenous parametrization (x^2,y^2,z^2,xy,xz,yz).</p>
<p><img src="http://mikael.johanssons.org/surfer/steiner.png" alt="Steiner's surface type 2 - immersion of the real projective plane." width="75%" /><br />
With the defining equation<br />
x^2*y^2-x^2*z^2+y^2*z^2-x*y*z=0<br />
this Steiner surface can be transformed into the Roman surface above if (and only if) you&#8217;re allowed to take shortcuts over the points at infinity. As it is, it has two pinches (both visible) and three lines of self-intersections (also all visible, kinda sorta). It&#8217;s also unbounded &#8211; one of the reasons that you cannot get to the bounded Roman surface easily.</p>
<p>With this as inspiration &#8211; go forth and draw surfaces. And when you do, please show them to me too.</p>
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		<title>The why and the what of homological algebra</title>
		<link>http://blog.mikael.johanssons.org/archive/2007/07/the-why-and-the-what-of-homological-algebra/</link>
		<comments>http://blog.mikael.johanssons.org/archive/2007/07/the-why-and-the-what-of-homological-algebra/#comments</comments>
		<pubDate>Thu, 12 Jul 2007 18:35:52 +0000</pubDate>
		<dc:creator>Michi</dc:creator>
				<category><![CDATA[Algebra]]></category>
		<category><![CDATA[Algebraic geometry]]></category>
		<category><![CDATA[Differential geometry]]></category>
		<category><![CDATA[Homology and Homotopy]]></category>
		<category><![CDATA[Mathematics]]></category>
		<category><![CDATA[Modular Representation Theory]]></category>

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		<description><![CDATA[I seem to have become the Goto-guy in this corner of the blogosphere for homological algebra. Our beloved Dr. Mathochist just gave me the task of taking care of any readers prematurely interested in it while telling us all just a tad too little for satisfaction about Khovanov homology. And I received a letter from [...]]]></description>
			<content:encoded><![CDATA[<p>I seem to have become the Goto-guy in this corner of the blogosphere for homological algebra. </p>
<p>Our beloved <a href=http://unapologetic.wordpress.com>Dr. Mathochist</a> just <a href=http://unapologetic.wordpress.com/2007/07/11/what-is-knot-homology/>gave me</a> the task of taking care of any readers prematurely interested in it while telling us all just a tad too little for satisfaction about Khovanov homology.</p>
<p>And I received a letter from the Haskellite crowd &#8211; more specifically from <a href=http://www.alpheccar.org>alpheccar</a>, who keeps on reading me writing about homological algebra, but doesn&#8217;t know where to begin with it, or why.</p>
<p>I <a href="http://blog.mikael.johanssons.org/archive/2006/01/introduction-to-algebraic-topology-and-related-topics-i/">have already</a> a <a href="http://blog.mikael.johanssons.org/archive/2006/02/monads-algebraic-topology-in-computation-and-john-baez/">few times</a> written about homological algebra, algebraic topology and <a href="http://blog.mikael.johanssons.org/archive/2006/05/my-first-group-cohomology-did-i-screw-up/">what it</a> <a href="http://blog.mikael.johanssons.org/archive/2006/11/a-for-the-layman/">is I do</a>, <a href="http://blog.mikael.johanssons.org/archive/2006/07/carry-bits-and-group-cohomology/">on various</a> <a href="http://blog.mikael.johanssons.org/archive/2006/07/triangulated-categories/">levels of difficulty</a>, but I guess &#8211; especially with <a href="http://carnivalofmathematics.wordpress.com">the carnival</a> dry-out I&#8217;ve been having &#8211; that it never hurts writing more about it, and even trying to get it so that the non-converts understand what&#8217;s so great about it.</p>
<p>So here goes.</p>
<p>Alpheccar writes that to his understanding, the idea is to build topological spaces out of algebraic gadgets, and then do topology on them. This is a part of the story, and certainly historically very important, but it is far from all of it.</p>
<h1>Motivations</h1>
<p>The revolution for homological algebra pretty much started with Eilenberg-MacLane &#8211; who wrote an <overline>article</overline>book that did the constructions necessary for the very topological versions of homological algebra &#8211; but without ever involving the actual topological spaces.</p>
<p>The point is that the way you do algebraic topology is that you tend to set up a functor Top &rarr; R-ChMod that assigns a chain complex of R-modules to each (nice enough) topological space, and then you add functors R-ChMod &rarr; R-Mod that extract informations from these. Typical examples are cellular chain complexes with coefficients somewhere nice for the first functor, and then homology or cohomology for the second functor &#8211; depending on what viewpoint is the most obvious.</p>
<p>The revolution was that we simply throw out that first functor. </p>
<p>In order to study (co)homology, we don&#8217;t really need to care that there was a topological space somewhere to begin with. We only, really, need a nice enough category of chain complexes (if it&#8217;s abelian, then that&#8217;s fine &#8211; we get the long exact sequences in homology and other niftiness easily then, but if it&#8217;s not, triangulated will do&#8230;) and we study certain types of functors from these to module categories.</p>
<h2>Homological algebra as a tool for algebraic topology</h2>
<p>Since, in the viewpoint introduced above, homological algebra is a part of the process used in algebraic topology, it turns out to be really neat to sit down and just prove a lot of neat results in homological algebra &#8211; with the background that at some later point, these might be useful once you sit down with the topology. I got hold of this particular point early &#8211; I had started my MSc thesis work in homological algebra before I took my first real topology course, and during that course, the less pointset topology and the more algebraic topology we did, the easier everything got. The fundamental results we needed to grasp to do algebraic topology in any amount of seriosity were basically just applications of all the cornerstone results in homological algebra, and thus perfectly obvious to my clique of arrogant undergrads.</p>
<p>This particular piece goes far. Why don&#8217;t we need to worry about whether we&#8217;re doing homology or cohomology? Answer: since Ext and Tor are dual in certain specific ways, which ends up meaning that although internal algebraic structures might be finicky, the module structure is very neat, and in k-Mod = Vect<sub>k</sub>, we end up with no worries anywhere.</p>
<h2>Testing grounds</h2>
<p>The viewpoint of homological algebra as a tool for algebraic topology goes pretty deep. When I ask my advisor what to put in texts where I motivate why our field is important, in the standard answer he gives me the following pops up:</p>
<blockquote><p>
Group cohomology is important, since it is a field where topological methods can be tested reasonably safely, since we have the group theoretic attack vectors in addition to the purely topological.
</p></blockquote>
<p>On the other hand, group cohomology also turns out to be important, since we get important information for the study of groups out of the homological algebra side of things.</p>
<h2>Low order Ext</h2>
<p>The area where this is most notable is in representation theory. This field comes in several flavours: group representations, where we study kG-Mod for some (sometimes finite) group G; Lie algebra representations, where we study g-Mod for some Lie algebra g; quiver representations, where we study kQ-Mod for some (finite) quiver algebra kQ &#8211; and so on. One question that tends to crop up here, and with a high degree of importance for the non-homological algebraists around me &#8211; is what happens if we know only parts of our group? Can we say something about the entire group based on that?</p>
<p>It turns out that we can. There are very neat correspondences between the lower order Ext groups over kG and the behaviour of G itself. I&#8217;m going to stick to group representations here, since that&#8217;s the area I know best &#8211; however, this is something that pops up analogously all over the place.</p>
<h3>Extensions</h3>
<p>Suppose you have some R-module K that you know embeds, in some specific way, into some larger R-module M. And suppose you find the quotient L=M/K in some manner. What could, then, M be? One obvious answer is <img src='/latexrender/pictures/fd114e6a3f85e9e6ef16868df610e399.png' title='G=K\oplus L' alt='G=K\oplus L' align='middle' />, but is this enough? This ends up depending on Ext<sup>1</sup><sub>R</sub>(L,K), with each element of this particular Ext group indexing precisely one such extension. </p>
<p>This is at the core of Maschke&#8217;s theorem, by the way, which says that if the characteristic of the field k doesn&#8217;t divide the group order |G|, then by a specific construction, the <b>only</b> extensions possible for <b>any</b> kG-modules are the split extension &#8211; the one where we just take the direct sum.</p>
<p>This all leads to a wealth of useful information and ideas in representation theory. For instance, there is a way to describe modules proposed by Dave Benson and some co-authors, where you draw diagrams with each vertex being a simple module, occupying that spot in a composition series, and the edges being taken from the relevant Ext<sup>1</sup>.</p>
<h3> Invariants and coinvariants</h3>
<p>Suppose you have a group acting on a vector space. This can be taken extremely physical &#8211; quantum mechanics is all about this kind of situation, or so I&#8217;ve heard. Then it might be interesting to figure out the invariant subspace: {a|ga=a for all g in G}. This is Ext<sup>0</sup>. Or we might want a basis for the complement: representatives for every way that things can move. This is the coinvariant vector space, defined as A/(ga-a), and this is just Tor<sup>0</sup>.</p>
<h3>Simples, projectives and the stable module category</h3>
<p>Simple modules are nice. They don&#8217;t have invariant subspaces. In the best of all worlds &#8211; which is when Maschke holds &#8211; simple modules are precisely the irreducible modules. However, when Maschke doesn&#8217;t hold, we can have non-trivial Ext<sup>1</sup>, and thus we can build larger modules out of simples by a kind of gluing: they aren&#8217;t just a nice direct sum of simples, but something ickier.</p>
<p>Thus, unless Maschke holds, there will be weird things happening in the module category.</p>
<p>These weird things, though, are controllable. To be specific, we can consider the smallest possible irreducible modules. These will end up being building blocks, and for nice enough worlds, these will also end up corresponding closely to the simples &#8211; in the way that we can allocate a simple to an irreducible projective in a bijective manner.</p>
<p>So &#8230; what <i>is</i> this projective I keep throwing around? Take a free module. This is a direct sum of a finite number of copies of the ring R. This will have direct summands. By picking apart all summands into further direct summands, at some point we hit bottom: we cannot pick anything apart any longer. This is, by the theorem of Krull-Schmidt, a well-defined state of being. We can permute things, but in essence, a module is just its decomposition into irreducibles.</p>
<p>So, anything that is a direct sum of a free module is a projective. We can lose projectivity by taking quotients &#8211; so if we add relations, we may well get lost. But as long as we just look for direct summands, we&#8217;re pretty much home free. Now, the irreducible projectives have to be summands of the ring R itself, so they end up actually being (left) ideals in the ring. And each of them corresponds intimately to a simple module.</p>
<p>One trick that&#8217;s very beloved among the people who worry about these things is to get rid of anything projective, and look at the stable module category. In this, we just quotient away anything projective &#8211; morphism sets are taken modulo morphisms that factor through a projective&#8230; This way, we only have the &#8220;essential&#8221;, or as it is known to the experts of the field &#8220;difficult&#8221; information left. Then Ext<sup>n</sup>(M,N)=Hom(&Omega;<sup>n</sup>M,N), where &Omega;<sup>n</sup> is the nth syzygy &#8211; see below for more on this.</p>
<p>So, homological algebra lets us understand the stable module category, which in turn lets us understand the parts that are essential to the module category structure.</p>
<h3>Resolutions</h3>
<p>I just promised you I&#8217;d tell you about syzygies. First off, some personal information &#8211; because readers always love that!</p>
<p>If you find me on IRC, on EFNet or on Freenode, you&#8217;ll find me under the nick Syzygy-. The &#8211; is there because there is someone who&#8217;s been using Syzygy for years and years on EFNet and because I&#8217;m not deliberately trying to be a bastard if I can help it. The rest of the nick is there to a certain extent because I like the way I write it in longhand.</p>
<p>And to a certain extent because it is an epitome of why homological algebra is interesting in my eyes.</p>
<p>Suppose we are interested in a finitely presented module, which we might be for many reasons, including being interested in algebraic geometry and in solving systems of polynomial equations. We might then just figure out what relations hold within a set of generators, which gives us a generating set, and some relation set.</p>
<p>These, relations, though are far from guaranteed to be the whole story. It&#8217;s probable that there are non-trivial relations between the relations. What do we do? We figure out what these are. They span the first <b>syzygy</b> module of the module we started with, denoted by &Omega;M. But this is unlikely to be free, so we can keep on going.</p>
<p>This way, we get a sequence of modules, all of which are free &#8211; since we just choose a generating set in each step &#8211; and with maps between them adding all the extra relations. But this is nothing other than a free resolution of the starting module. And here comes the candy that hooked me for this discipline: studying modules over their resolutions is the same thing as studying what chain complexes are, deep down, which in turn is the same thing as studying homological algebra.</p>
<p>Want to figure out what a module map means for the family of syzygies? What you really want is a chain map in the chain complex category. But some of these maps &#8211; or even portions of maps &#8211; will not carry relevant information. So we factor those away, and we get a slightly weirder category. But here, equality doesn&#8217;t quite mean what it should, so we add in more equality relations. And suddenly, we live in a derived category &#8211; and in here, the Hom sets are Ext groups, and the tensor products are Tor groups.</p>
<h2>Number theory, geometry, and computation!</h2>
<p>To continue this tour de force, consider the theorems in vector calculus relating various triple and double integrals. (note &#8211; I never dealt with this. I rode on technicalities to root out calculus from my curriculum so it would fit more algebra&#8230;.) These theorems, in the end, only state that <img src='/latexrender/pictures/55848c4b2b9c6041b11c02125da13392.png' title='\partial^2=0' alt='\partial^2=0' align='middle' />, which is at the core of what homological algebra is all about.</p>
<p>If we formalize this particular recognition a bit, and tug at the corners, we end up with de Rham cohomology, which deals with what you can do with differential forms on manifolds (layman speak: things you can integrate. The f(x)dx after the integral sign is a typical differential form) &#8211; and this is one of the many many places where cohomology rather than homology ends up being the &#8220;right&#8221; way to view things just because you started out as a geometer instead of .. well .. topologist or algebraist.</p>
<p>The same kind of thing happens in algebraic geometry as well. You start out happily with your varieties, you conclude that as soon as things get interesting, the nice and pretty concepts of coordinate rings don&#8217;t hold up, and you&#8217;re forced to go to coordinate sheaves. And then you try to figure out what you can do with sheaves of functions on a variety &#8211; and before you know it, you reconstructed sheaf cohomology. This, by the way, a quick look at wikipedia told me, lets you define euler characteristics for varieties in a way consistent with the classical uses of it.</p>
<p>I am no geometer, and I&#8217;m not the person to tell you about the intricacies of these things. If you understand them, though, I&#8217;d love to figure them out at some point.</p>
<p>The discussion of Khovanov homology is a slightly (though not very) similar thing to this. Again, I have no real idea, and am treading on thin ice here.</p>
<p>So, alpheccar. Is this what you asked for? Please tell me what more you want covered, and I&#8217;ll write up some more! This was fun writing!</p>
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