2
$\begingroup$

Prove that $\mu:k^n\rightarrow \text{maximal ideal}\in k[x_1,\ldots,x_n]$ by $$(a_1,\ldots,a_n)\rightarrow (x_1-a_1,\ldots,x_n-a_n)$$ is an injection, and given an example of a field $k$ for which $\mu$ is not a surjection.

The first part is clear, but the second part needs a field $k$ such that not all maximal ideals of the polynomial ring is of the form $(x-a_1,\ldots,x-a_n)$. I am not sure how to find one as I obviously need to a non-obvious ring epimorphism $k[x_1,\ldots,x_n]\rightarrow k$ such that the kernel is the maximal ideal. This question is quite elementary and I feel embarrassed to ask.

$\endgroup$
  • $\begingroup$ What does $\text{maximal ideal} \in k[x_1,\ldots,x_n]$ mean? Oh it's the set of maximal ideals in $k[x_1,\ldots,x_n]$, I get it. And $(x_1-a1,\ldots,x_n-a_n)$ is not a tuple, but the ideal generated by it's entries! (Sorry.) $\endgroup$ – k.stm Nov 4 '12 at 20:09
  • $\begingroup$ So, you need to find a maximal ideal of a polynomial ring whose field is not algebraically closed. This is a consequence of Hilbert's weak Nullstellensatz. $\endgroup$ – Rankeya Nov 4 '12 at 20:10
  • 3
    $\begingroup$ Try $k=\mathbb R, n=1$ $\endgroup$ – Georges Elencwajg Nov 4 '12 at 20:10
  • 2
    $\begingroup$ To elaborate a little on @GeorgesElencwajg comment: $\mathbb{R}[x]$ is a PID. So, all non-zero prime ideals are maximal. But, $\mathbb{R}[x]$ has irreducible polynomials other than degree 1 polynomials. $\endgroup$ – Rankeya Nov 4 '12 at 20:12
  • 2
    $\begingroup$ Dear @Julian, I have done what you requested. $\endgroup$ – Georges Elencwajg Sep 12 '13 at 23:09
5
$\begingroup$

At Julian's request I'm developing my old comment into an answer. Here is the result:

Given any non algebraically field $k$, the canonical map $$k\to \operatorname {Specmax}(k[x]):a\mapsto (x-a)$$ is not surjective.

Indeed, by hypothesis there exists an irreducible polynomial $p(x)\in k[x]$ of degree $\gt 1$.
This polynomial generates a maximal ideal $\mathfrak m=(p(x))$ which is not of the form (x-a), in other words which is not in the image of our displayed canonical map.

$\endgroup$

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.