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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.

Michael Hardy
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Bombyx mori
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  • 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.) – k.stm Nov 04 '12 at 20:09
  • 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. – Rankeya Nov 04 '12 at 20:10
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    Try $k=\mathbb R, n=1$ – Georges Elencwajg Nov 04 '12 at 20:10
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    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. – Rankeya Nov 04 '12 at 20:12
  • Thanks for the help! I clearly lost my mind. – Bombyx mori Nov 04 '12 at 20:21
  • @GeorgesElencwajg Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Sep 12 '13 at 22:47
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    Dear @Julian, I have done what you requested. – Georges Elencwajg Sep 12 '13 at 23:09

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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.

Georges Elencwajg
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