# Localisation and ring depth

So, we're back at the point where I'm hesitating whether what I tried to work out even made sense or not. So I'll do a write up of all the things I feel certain about asserting, and ask my loyal readership to hunt my errors for me. :)

Don't laugh. This is less embarrassing for me than asking my advisor point blank.

We say that a (graded) commutative ring R has depth k if we can find a sequence of elements [tex]x_1,\dots,x_k[/tex] with [tex]x_1[/tex] not a zero-divisor, each [tex]x_i[/tex] not a zero-divisor in the quotient [tex]R/\langle x_1,\dots,x_{i-1}\rangle[/tex] and [tex]R/\langle x_1,\dots,x_k\rangle[/tex] a ring without non-zero divisors. This definition, of course, being the first obvious point where I may have screwed up.

Now, we know (from looking it up in Atiyah-MacDonald), that for SR the localisation of R in a multiplicatively closed subset S, S(R/I)=SR/SI, that injections carry over to injections, and that the annihilator over SR of an element is the localisation of the annihilator of the element.

We now take S=R-p for p a prime ideal in R. So SR is the localisation in a prime ideal in the usual sense (by slight abuse of notation). Since p doesn't intersect S, Sp is prime in SR. Suppose we have an R-sequence of length k, demonstrating the depth of R to be at least k.

The first element, [tex]x_1[/tex] of the R-sequence, is regular in R, and thus multiplication by this element is an injection [tex]R\to R[/tex]. Injections carry over to injections after localisation, so this induces an injection [tex]SR\to SR[/tex], given by [tex]\frac st\mapsto \frac{x_1s}t[/tex]. Since this is injective, the element [tex]\frac{x_1}1[/tex] is regular in SR. So we can go to [tex]SR/\langle\frac{x_1}1\rangle[/tex]. Now, localisation commutes with quotients, so this is really [tex]S(R/\langle x_1\rangle)[/tex].

Now, [tex]x_2[/tex] is regular in [tex]R/\langle x_1\rangle[/tex], and thus induces an injective map. This injective map carries over to an injective map of localisations, and thus, with the same argument as above, [tex]\frac{x_2}1[/tex] is regular over [tex]S(R/\langle x_1\rangle)[/tex]. So we can carry over all of our regular sequence from R to SR.

The only problem visible to me being that we may end up dividing out the entire ring and ending up with a trivial ring earlier than we thought we would. What would make that happen? Well, if our ideal in the quotient at some point includes the ring, we lost the game. What could make this happen? Then we'd need to have either 1 in the ideal, or 0 in the localisation set. 0 in the localisation set, we get if a zerodivisor and its annihilator both end up in there at the same time. I'm not going to touch this case quite yet. Possibly not at all in this post. 1 in the ideal, we get if one of the [tex]x_i[/tex] ends up being in S. So, for everything to be perfectly fine this far, we REALLY want to avoid straying outside of p.

And this'd, y'know, limit our choices of R-sequences a bit. But still, we can get an R-sequence the depth of SR this way, I think. And then there is the question of what happens outside this realm - and it would seem that we can use the Krull dimension bound for that, thus arriving at

[tex]\operator{depth} R \leq \operator{depth} SR + \dim R/p[/tex]

(which incidentially, with a certain set of conditions, is a theorem I found in a set of notes that sqrt linked me to)