Posted by: hilbertthm90 | November 9, 2009

Some Corollaries

Today will just be some quick results we get from this build up.

First, if we localize a polynomial ring at a maximal ideal, say k[x_1, \ldots, x_n] at \frak{m}=(x_1, \ldots, x_n), then \dim R_\frak{m}=n=\dim R. This is because G_m(R) has Poincare series (1-t)^{-n} so the order of the pole is n which is the dimension by the last post.

This one will be really useful later: \dim R\leq \dim_k(\frak{m}/\frak{m}^2). Let \{x_i\}_1^r \subset\frak{m} such that \overline{x_i}\in \frak{m}/\frak{m}^2 are a basis for the vector space. Then by Nakayama’s Lemma the x_i generate \frak{m}. Thus \dim_k(\frak{m}/\frak{m}^2)=s\geq \dim R.

This one is also useful in algebraic geometry. If R is Noetherian, and x_1, \ldots , x_r\in R, then every minimal ideal \frak{p} belonging to (x_1, \ldots, x_r) has height \leq r. Unfortunately, we cannot push this to equality. Geometrically the example is that if Y is the twisted cubic, then I(Y) has height 2, but cannot be generated by less than 3 elements.

Lastly, we’ll do the famous Principal Ideal Theorem. If R is Noetherian and x\in R is neither a zero-divisor nor a unit, then every minimal prime ideal \frak{p} of (x) has height 1. By the last paragraph we know that ht(p)\leq 1. If ht(\frak{p})=0 then it belongs to (0). Thus every element of \frak{p} is a zero-divisor which is a contradiction since x\in \frak{p}.

Posted in algebra | Tags: , ,

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