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In Reid's Undergraduate Commutative Algebra, $k$ a field and a point $P=(a_1,\dots,a_n)\in k^n$ determine a homomorphism on the the polynomial ring of functions $k[x_1,\dots,x_n]\to k$ by $g\mapsto g(P)$ for $g\in k[x_1,\dots,x_n]$.

The kernel of this homomorphism is $(x_1-a_1,\dots,x_n-a_n)$. Reid denotes this as $m_P$ and says it is maximal. Why is it maximal though? I know $k[x_1,\dots,x_n]/(x_1-a_1,\dots,x_n-a_n)$ is isomorphic to the image of the map in $k$, which is a commutative subring of a field. Is there a way to see it is a field, to get that the kernel is maximal? I figure if $g(P)\neq 0$ in $k$, then is $1/g(P)$ the value of some other polynomial $f(P)$?

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2 Answers 2

up vote 5 down vote accepted

The map $k[x_1, \ldots, x_n] \to k$ that you wrote down is surjective, since for each $b \in k$ we can always take $g$ to be the constant polynomial $g(x_1, \ldots, x_n) = b$, for which $g(P) = b$.

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Thanks, I get it now. –  hmIII Mar 6 '12 at 1:51

Because the map is surjective and has codomain a field.

One of the basic theorems tells you that, since the map is surjective, the ideals of the codomain are in bijection with the ideals of the domain which contain the kernel. Since the codomain has exactly one proper ideal, that means there is one proper ideal in the domain of the map containing its kernel, namely, the kernel itself.

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I in fact always define a maximal ideal to be one such that the corresponsing quotient of the ring is a field. This is almost always a more practical description than the usual "maximal among the proper ideals". The argument you give then serve to obtain that other description, and justify the name "maximal", Similalry prime ideals are those with their quotient an integral domain, and one gets "maximal implies prime" for free. –  Marc van Leeuwen Mar 5 '12 at 4:55

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