# Maximal ideals in $K[X_1,\dots,X_n]$

Let $K$ be a field, and $a_1,\dots,a_n \in K$. Prove that the ideal $$(X_1-a_1,\dots,X_n-a_n)$$ is maximal in $K[X_1,\dots,X_n]$.

I tried proving that the only elements outside the ideal are the invertibles of $K$ (I should still prove that this implies maximality, but it shouldn't be too difficult).

Is there a better strategy, or another stategy?

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It may be interesting to you to note that the opposite direction is also true, i.e.: any maximal ideal in that multivariable polynomial ring is of the given form, yet any proof of this other direction I know uses heavily analytic tools, whereas the direction you're asking is purely algebraic. – DonAntonio Jan 22 '13 at 16:18
It's easier to see that the ideal $(X_1, \ldots, X_n)$ is maximal and your question's like a "change of variable". DonAntonio, can you say me where you saw the proof of the opposite direction? – Diego Silvera Jan 23 '13 at 1:27
@DonAntonio: Dear Don, The converse is true only if $K$ is algebraically closed. (Think about the case $n = 1$.) Regards, – Matt E Jan 24 '13 at 5:25
Good catch, @MattE. Thanks. – DonAntonio Jan 24 '13 at 5:39
Several answers provide strategies that work, but it may be useful to also point out that the strategy you suggested in the question, "proving that the only elements outside the ideal are the invertibles of $K$," is doomed to failure because that isn't true. The elements outside the ideal are all the polynomials $f$ such that $f(a_1,\dots,a_n)\neq0$, and that includes lots of non-constant polynomials (i.e., polynomials that aren't in $K$). – Andreas Blass Jan 24 '13 at 14:19

Hint: Define

$$f:K[X_1,...,X_n]\to K\;\;,\;\;f(g(X_1,...,X_n)):=g(a_1,...,a_n)$$

1) Show $\,f\,$ is a surjective ring homomorphism

2) Use now the first isomorphism theorem for rings

3) Remember: if $\,R\,$ is a commutative unitary ring, an ideal $\,I\leq R\,$ is maximal iff $\,R/I\,$ is a field.

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Let $f_1$ be the homomorphism $$f_1:K[X_1,\cdots,X_{n-1}][X_n]\rightarrow K[X_1,\cdots,X_{n-1}]$$ $$f_2:K[X_1,\cdots,X_{n-2}][X_{n-1}]\rightarrow K[X_1,\cdots,X_{n-2}]$$ $$\cdots$$ $$f_n:K[X_1]\rightarrow K$$ Let assume that we know each $f_k$ to be a ring homomorphism. Then the composition $$f=f_n \cdots f_1$$ is an homomorphism. It is surjective because for each $a\in K$ we can evaluate the constant polynomial $a$. The kernel is $N:=\{g\in K:g(a_1,\cdots,a_n)=0\}$ I should now proof that $N=(X_1-a_1\cdots X_n-a_n)$ but is Ruffini still valid? – Temitope.A Jan 24 '13 at 22:02

Hint $\ \ (I,f) = (I,f\ mod\ I) = (I,f(\bar a))\,\ [\,= 1 \iff f(\bar a)\ne 0\iff f\not\in I]$

Remark $\$ It is instructive to compare this internal approach to the structural approach mentioned by DonAntonio.

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@YACP $\ I = (\bar X- \bar a) = (X_1-a_1,\ldots,X_n-a_n),\$ and $\, {\rm mod}\ I\!:\ X_i\equiv a_i\Rightarrow f(X_1,\ldots,X_n) \equiv f(a_1,\ldots,a_n) = f(\bar a)\ \$ – Math Gems Jan 24 '13 at 15:52
@YACP Ideals are always preserved under the operation of "modding out" one generator by others, i.e. $(a_1,\ldots,a_n,b) =$ $(a_1,\ldots,a_n,b\!-\!r_1 a_1-r_2 a_2\! -\!\cdots- r_n a_n).\,$ Above the modular reduction corresponds to the multidimensional generalization of the polynomial remainder theorem. – Math Gems Jan 24 '13 at 16:24
@YACP Even though, in general, there is no (generalized) division with (unique) remainder (with associated "mod" operation), it is conceptually helpful to think of such ideal tranformations like the mod operations employed in the reduction steps of the Euclidean algorithm (or the multidimensional generalizations used in standard/Grobner basis reductions). – Math Gems Jan 24 '13 at 16:24
What do you mean by "multidimensional generalization of the polynomial remainder theorem"? As far as I can see you use the following result: $f$ can be written as a combination of $X_i-a_i$ (with polynomial coefficients) plus an element of $K$. Is this obvious or deserves a proof? – user26857 Jan 24 '13 at 16:32
@YACP It's an easy inductive proof: apply the univariate remainder theorem to the highest variable, using $k[X_1,\ldots,X_n] = R[X_n]\:$ for $\:R = k[X_1,\ldots,X_{n-1}].\ \$ – Math Gems Jan 24 '13 at 16:51

Let $P(X_1, \ldots, X_n)$ a polynomial. Substitute $X_i\mapsto X_i + a_i$ and get $$P(X_1+ a_1, \ldots, X_n + a_n) = \sum c_{\alpha} X_1^{\alpha_1} \ldots X_n^{\alpha_n}$$ and so

$$P(X_1, \ldots, X_n) = \sum c_{\alpha} (X_1-a_1)^{\alpha_1} \ldots (X_n-a_n)^{\alpha_n}$$

Note that $c_{(0,\ldots, 0)} = P(a_1, \ldots, a_n)$. Moreover, $$P(X_1, \ldots, X_n) = c_{(0,\ldots, 0)}+ \sum (X_i - a_i) g_i(X_1, \ldots, X_n)$$ as all the other terms $c_{\alpha}(X-a)^{\alpha}$ are divisible by some $(X_i - a_i)$. Therefore $$P(X_1, \ldots, X_n) - P(a_1, \ldots, a_n) \in (X_1-a_1, \ldots, X_n - a_n)$$

and therefore $P(a_1, \ldots, a_n) \in (P, (X_1 - a_1) , \ldots, (X_n - a_n))$. Assume moreover that $P \not \in (X_1- a_1, \ldots X_n - a_n)$. Then $P(a_1, \ldots, a_n) \ne 0$ and we conclude that $1 = P(a_1, \ldots, a_n)^{-1} \cdot P(a_1, \ldots, a_n) \in (P, (X_1 - a_1), \ldots, X_n - a_n)$. Therefore $(X_1-a_1, \ldots, X_n - a_n)$ is maximal.

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I realize that we should avoid responding to other answers, but when they make false statements there should be a way to correct them.

$\mathbb Q$ is a field. ${X_1}^2-2$ is a prime element in $\mathbb Q\left[ X_1\right]$ which is a p.i.d so ${X_1}^2-2$ generates a maximal ideal. The converse of this statement is false for the field $\mathbb Q$ or any field that is not algebraically closed. If $K$ is algebraically closed, both the statement of the question and its converse are corollaries of the Hilbert Nullstellensatz. This basic result in algebraic geometry can be found in texts on algebraic geometry, for example Eisenbud's Commutative Algebra with a view toward algebraic geometry, Springer Graduate Texts in Math, vol 150, pp 34--35.

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Dear Barbara, The comment about the converse was made in a comment, and you can correct it by making a comment in reply (as I did above). Regards, – Matt E Jan 24 '13 at 5:27
Hi Barbara, I am not sure how it happened, but your account split, despite you logging into the system using the same credentials. If you again find yourself unable to post comments or edit your own posts without review, please let a moderator know so we can track down the problem. I've merged your two accounts now. – Willie Wong Jan 24 '13 at 15:59
Thank you very much for your help. Things seem to be working fine now, and I am commenting appropriately now. – Barbara Osofsky Jan 25 '13 at 19:19