1

For the past couple of days i had been working on an interesting homework problem, my interpretation of which is as follows:

Let $m$ be a maximal ideal of $k[x_1,\cdots,x_n]$. Let $m_i = m \cap k[x_i], i=1,\cdots,n$. Then show that $m$ is the unique pre-image of $m_1,\cdots,m_n$, under the map $\operatorname{Specm} k[x_1,\cdots,x_n] \rightarrow \prod_i \operatorname{Specm} k[x_i]$ given by $m \mapsto (m_i)_i$, if and only if $\left[k[x_1,\cdots,x_n]/m:k\right] = \prod_i \left[k[x_i]/m_i :k\right]$.

Related to this problem are the two questions of mine: Intersecting maximal ideals of $k[x_1,\dots,x_n]$ with $k[x_i]$ and Conditions for the degree of field extension reaches its upper bound.

PS: My homework is submitted, please feel free to attack to the problem :)

Community
  • 1
Manos
  • 26,949

1 Answers1

1

There exists the following counter-example for your assertion.

Suppose $char(k) = p > 0$. Let $\bar k$ be an algebraic closure. Let $a \in k$. Suppose $x^p - a$ is irreducible in $k[x]$. Let $m_1 = (x_1^p - a)$, $m_2 = (x_2^p - a)$. $x^p - a$ has a unique root $\alpha$ in $\bar k$. Hence there exists unique maximal ideal $m$ of $k[x_1, x_2]$ such that $m_i = m \cap k[x_i]$ fo $i = 1, 2$. Since $[k(\alpha) : k] = p$, $[k[x_1, x_2]/m : k] = p$ and $[k[x_i]/m_i: k] = p$ for $i = 1, 2$.

Makoto Kato
  • 44,216
  • Thanks for your answer. I am currently studying it. I think there is a typo: "$x^p-\alpha$ has a unique root $\alpha$ in $\bar{k}$". – Manos Oct 15 '12 at 18:00
  • @Manos It's not typo. It is true that $x^p - a$ has a unique root in $\bar k$. so does $x_1^p - a$ or $x_2^p - a$. – Makoto Kato Oct 15 '12 at 18:06
  • I agree that $x^p-\alpha$ has a unique root, but is this root equal to $\alpha$? In other words, why do we have that $\alpha^p=\alpha$? That is my point. – Manos Oct 15 '12 at 18:09
  • @Manos I did not write $x^p - \alpha$. I wrote $x^p - a$. – Makoto Kato Oct 15 '12 at 18:14
  • I understand. In your argument, is it true that $x_1 = x_2 , , (mod , , m)?$ – Manos Oct 15 '12 at 18:32
  • @Manos Yes. There exists a unique homomorphism $f\colon k[x_1, x_2] \rightarrow \bar k$ such that $f(x_1) = \alpha$ and $f(x_2) = \alpha$. $m = Ker(f)$. Since $f(x_1 - x_2) = 0$, $x_1 \equiv x_2$ (mod $m$). – Makoto Kato Oct 15 '12 at 19:13
  • Excellent as always :) Thanks. – Manos Oct 15 '12 at 19:15