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Let $A$ be the polynomial $K$-algebra $K[t_1, t_2]$. Then the algebra $A$ is not local, why? More generally, $K[t_1, t_2, \cdots, t_n]$ is not local for any $n\geq 1$, why?

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up vote 1 down vote accepted

How about the fact that you can find two distinct maximal ideals of $K[t_1, t_2]$? Like $(t_1,t_2)$ and $(t_1 - 1, t_2 - 1)$, for instance. In general, $K[t_1, \dots, t_n]$ is not local because of the ideals $$ (t_1, \dots, t_n) \quad \text{ and } (t_1 - 1, \dots, t_n - 1). $$ Hope that helps,

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You are right, thank you very much Patrick Da Silva – Aimin Xu Nov 3 '12 at 2:13
  • If $A$ is local, then there does not exist a non-unit $f$ such that $f-1$ is also a non-unit.
  • If $A$ is local, then it does not have two maximal ideals.
  • If $A$ is local, then there can be at most one homomorphism $A \to K$ that is the identity on $K \subseteq A$.
  • ...

If you can manage to show any one of these conclusions false, you've proven $A$ is not local.

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(If it's not clear where the third bullet point comes from, it' can be derived as a special case of a reformulation of the second bullet point) – Hurkyl Nov 3 '12 at 0:09

Try constructing a homomorphism from $K[x_1, ..., x_n]$ into $K$ whose kernel is $(x_1 - k_1, ..., x_n -k_n).$ Since $K$ is a field, what can we say about this ideal?

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Seems like the easiest way to look at all of them is this way: $t_1$ and $1-t_1$ aren't units, so they generate proper ideals $(t_1)$ and $(1-t_1)$. If there were a unique maximal proper ideal containing both, it would have to contain $t_1+(1-t_1)=1$, but that would contradict the fact the ideal is supposed to be proper.

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+1: This elegant proof shows that given any (commutative) ring $R$, the ring $A=R[t]$ is never local. – Georges Elencwajg Nov 3 '12 at 9:16

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