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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.

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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 4 '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 4 '12 at 20:10
Try $k=\mathbb R, n=1$ – Georges Elencwajg Nov 4 '12 at 20:10
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 4 '12 at 20:12
Dear @Julian, I have done what you requested. – Georges Elencwajg Sep 12 '13 at 23:09
up vote 5 down vote accepted

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.

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