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)