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I want to use the result that the polynomial ring $R:=k[x_1,\cdots,x_n]$ is a regular ring.

I can prove that every maximal ideal $\mathfrak m$ in $R$ can be generated by $n$ number of elements so that $R_{\mathfrak m}$ is a regular local ring.

Now Serre's result shows that if $A$ is a regular local ring then $A_{\mathfrak p}$ is a regular local ring. From this it follows that $R$ is a regular ring. Now Serre's result is a BIG theorem, so is there any other easier way to prove this result. What if we assume $k$ is algebraically closed?

I have a presentation $+$ viva (on some topic in commutative algebra) coming up next week and I need to use the above result. So what I kind of question I can expect when I use this theorem?

I know I have asked many questions in one post but these are all related. I am sorry for this.

Thank you.

  • Can you give a proof that every maximal ideal of R can be generated by n elements? And please explain why $R_m$ is regular local ring by this way. Thank you. – Jian Jan 20 '18 at 16:00
  • @Sky If we prove that $\mathfrak m$ can be generated by $n$ elements say ${f_1,\cdots , f_n}$ then $\mathfrak m R_{\mathfrak m}$ can be generated by $n$ elements and $\dim R_{\mathfrak m}=\operatorname{ht}\mathfrak m=\dim R=n$, so that $R_{\mathfrak m}$ is regular. And every maximal in $R$ can be generated by $n$ elements, this statement is proved in for example Matsumura Commutative ring theory theorem $5.3$, page $33$. –  Jan 20 '18 at 16:20
  • Just to be clear, are you asking whether one can give a direct proof that $PR_P$ is generated by $\dim R_P$ elements for any prime ideal $P\subset R$, $R$ a polynomial ring over a field? – Mohan Jan 20 '18 at 20:04
  • @puasam you should prove the minimal number of generators equal to the krull dimension of $R_m$.I don't think it is easy as you have said. – Jian Jan 21 '18 at 00:53
  • @Mohan Yes my question is exactly that. May be if we can assume $k$ is algebraically cllosed? –  Jan 21 '18 at 03:01
  • @Sky $\mathfrak mR_{\mathfrak m}={\frac{f_1}{1},\cdots ,\frac{f_n}{1}}$. And for a polynomial ring every maximal ideal has height equal to its dimension. I am not saying any proof is easy. –  Jan 21 '18 at 03:05

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This can be proved if you know Noether Normalization, which essentially uses the fact any non-constant polynomial in $n$ variables can be made monic in one of the variables after a change of variables. So, let us assume this.

Then, maximal ideals are $n$ generated follows by induction on $n$ as follows. $n=1$ being trivial, assume proved for $n-1$ and let $M\subset R=k[x_1,\ldots, x_n]$ be a maximal ideal. Then, after change of variables, we may assume (since $M\neq 0$), $M$ contains a monic polynomial in $x_n$. Let $N=M\cap S$ where $S=k[x_1,\ldots, x_{n-1}]$. We have thus an integral extension $S/N\to R/M$, but $R/M$ is a field implies by integrality, that so is $S/N$. So, $N$ is a maximal ideal of $S$ and so by induction generated by $n-1$ elements. Then, the image of $M$ in $S/N[x_n]$ is a maximal ideal and so by one variable case, we get that it is one generated, and thus $M$ itself is $n$-generated.

For a prime ideal $P$, we can by Noether Normalization, assume that $k[x_1,\ldots,x_k]\subset R/P$ is an integral extension where height of $P=n-k$. This says, in particular, $P\cap k[x_1,\ldots,x_k]=0$ and thus $R_P$ is a localisation of $k(x_1,\ldots,x_k)[x_{k+1},\ldots,x_n]$ at the image of $P$. The integrality says the image of $P$ in this ring is a maximal ideal and thus, by the previous part, generated by $n-k$ elements which implies, so is the localization. Hope rest is clear.

Mohan
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  • @Mohan Can you please explain what you did after you proved $N$ is generated by $n-1$ elements. I am kind of lost after that. –  Jan 21 '18 at 04:31
  • @puasam I showed above that $N$ is $n-1$ generated over $S$ and thus so is $NR$ over $R$. $NR\subset M$ and I also showed that $M/NR$ is one generated over $R$. Together, we see that $M$ is $n$-generated over $R$. – Mohan Jan 21 '18 at 20:19