7
$\begingroup$

Let $(R,\mathbb{m})$ be a Noetherian local ring and $I$ an ideal of $R$. Let $t$ be an indeterminate over $R$. The analytic spread $l(I)$ of $I$ is defined to be the Krull dimension of the ring $R[It]/\mathbb{m}R[It]$. Let $x$ be another indeterminate over $R$, then $\mathbb{m}R[x]$ is a prime ideal of $R[x]$ and $R\rightarrow R[x]_{mR[x]}$ is a faithfully flat ring extension, Let $R(x)=R[x]_{\mathbb{m}R[x]}$.

The question is: Why $l(I)=l(IR(x))$?

This is a conclusion in Lemma 8.4.2 in I. Swanson and C. Huneke's book " Integral closure of ideals, rings and modules". I don't know how to prove it, any help will be appreciated!

$\endgroup$
0
4
$\begingroup$

Let $A=R[It]/\mathfrak{m}R[It]$. Let $k=R/\mathfrak{m}$. Let $B=R[x]_{\mathfrak{m}R[x]}$, $J=IB$, $C=B[Jt]/\mathfrak{m}B[Jt]$. We need to show that the Krull dimensions of $A$ and $C$ are same.

Then $B/\mathfrak{m}B=k(x)$, $A=R/\mathfrak{m}\otimes_R R[It]$ and $C=B/\mathfrak{m}B\otimes_BB[Jt]$.

Since $B$ is flat over $R$, we have $B\otimes_RI^n=I^nB=J^n$ for any $n\geq 0$, it follows that $B[Jt]=B\otimes_RR[It]$ as $R$-modules, but the isomorphism is also a ring map (check it), so $B[Jt]=B\otimes_RR[It]$ as rings.

Now $C=B/\mathfrak{m}B\otimes_B(B\otimes_RR[It])=B/\mathfrak{m}B\otimes_RR[It]=(B/\mathfrak{m}B\otimes_{R/\mathfrak{m}}R/\mathfrak{m})\otimes_RR[It]=k(x)\otimes_kA$.

We only need to show $k(x)\otimes_kA$ and $A$ have the same Krull dimension.

Notice that $R$ is Noetherian, $I$ is finitely generated, say $I=(x_1,\ldots,x_n)$, then $A=k[\overline{x}_1t,\ldots,\overline{x}_nt]$ where $\overline{x}_i$ denotes the image of $x_i$ in $I/\mathfrak{m}I$, thus $A$ is a finitely generated $k$-algebra.

By Noether's normalization lemma, we can find algebraically independent elements $y_1,\ldots,y_r\in A$ such that $A$ is integral over $k[y_1,\ldots,y_r]$. So $k(x)\otimes_kA$ is integral over $k(x)\otimes_kk[y_1,\ldots,y_r]=k(x)[y_1,\ldots,y_r]$, thus $k(x)\otimes_kA$ is of dimension $r$ ($=\dim A$).

We win.

$\endgroup$
1
  • $\begingroup$ user119882, thank you! To use Noether's normalization lemma is necessary which I did not know, thank you! $\endgroup$ – gaotian81 May 8 '14 at 13:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.