Michi’s blog » Localisation and ring depth

## Localisation and ring depth

• September 5th, 2006
• 1:24 pm

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 with not a zero-divisor, each not a zero-divisor in the quotient and 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, of the R-sequence, is regular in R, and thus multiplication by this element is an injection . Injections carry over to injections after localisation, so this induces an injection , given by . Since this is injective, the element is regular in SR. So we can go to . Now, localisation commutes with quotients, so this is really .

Now, is regular in , 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, is regular over . 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 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

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

### 8 People had this to say...

• Craig
• September 6th, 2006
• 8:31

Hi Michi, does the “estimated 2.12 minutes” reading time when you click to continue reading the rest of your post take into account the time needed to understand the maths therein?

Perhaps readers should register and complete a short test to assess their mathematical knowledge and then you could calculate more accurately…

• Michi
• September 6th, 2006
• 9:34

Nah, I’d rather just find out how to turn the thing off instead then. It’s an artifact of the preview plugin I’m using.

How much time DID you use to grok what I was writing?

• folk angel
• September 6th, 2006
• 14:04

Well, it’s not much use to you, but it took me 10 mins to get my head around your definition of the depth of a (graded) commutative ring* – it sounds plausible enough but as your uncertain as to its accuracy where did you get it from? (presumably not Atiyah-MacDonald – and I can’t find it to verify in my limited remaining supply of scary yellow maths books (Eisenbud being the most relevant looking) or by my usual cheat of looking it up on mathworld).

Thanks for waking my brain up though and good luck with the Phud.

*then it is 3 years since I did any maths and whilst I did used to know some things about commutative rings, I don’t recall depth being one of them.

• Michi
• September 6th, 2006
• 14:18

Atiyah-MacDonald doesn’t treat depth at all (I’ve been staring at it and swearing while working with this), my Eisenbud is in Jena – some 8 hours by train from here, and the local maths library doesn’t carry anything that helped me.

The definition is basically what I boiled down from how I think about it, and how I remember it being defined. I am certain of the formulation in terms of length of longest sequence of sequential non-zero divisors, but not certain as to what extent my formulation captured all possible subtleties.

There is another formulation of it in terms of vanishing of Ext-modules – basically, the depth (when finite) is the degree of the first non-vanishing Ext. Had I had a decent reference while writing this, my uncertainty would have been nowhere near the currently exhibited levels.

• Michi
• September 6th, 2006
• 14:19

(administrative note: Since it doesn’t accurately reflect reading times anyway, I have now removed that particular bit)

• folk angel
• September 6th, 2006
• 14:40

have failed to find a definition of depth of graded ring on internut, only for algebras:

http://planetmath.org/encyclopedia/HomogeneousSystemOfParameters.html

or modules:

http://eom.springer.de/d/d031180.htm

but my brain is too atrophied these days to work out if they’re relevant.

• Michi
• September 6th, 2006
• 14:53

Well, they are kinda. A graded ring is – in the cases I’m interested in – always a graded algebra over some field. Thus, the regular sequences exhibiting lower bounds on depth are all parts of homogenous systems of parameters, but it may happen that some of the parameters occuring end up being zero-divisors, so it may happen that not all of them occur.

The “usual” definitions tend to be for the depth of a module over a specific ring, with the more or less tacit understanding that the ring as such has depth given as the depth of the ring viewed as the obvious module over itself.

• Craig
• September 14th, 2006
• 16:17

Sorry, I got sidetracked and forgot to keep up with the comments… I must confess that I didn’t actually read the maths in too much detail as I was having a bad-math-week and I couldn’t bring myself to do any extra!