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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?

glS
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Temitope.A
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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
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    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 01:27
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    @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 05:25
  • Good catch, @MattE. Thanks. – DonAntonio Jan 24 '13 at 05:39
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    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
  • @AndreasBlass What do you mean by "polynomials that aren't in K"? Do you mean the roots? – Temitope.A Jan 24 '13 at 21:37
  • to @DonAntonio coment the convers can be prove with algebraic tools to , is the nullstellenstaz weak theorem (that is not weak at all) i am learning it in a introduccion curse to algebraic geometry. – Fran Aguayo Apr 09 '23 at 04:38

6 Answers6

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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.

DonAntonio
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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
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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.

orangeskid
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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.

Math Gems
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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
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    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? –  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
  • @YACP That's why I encouraged the reader to compare it to the external "structural" approach. This will help one to understand how the multivariate polynomial remainder theorem works. Moreover, it will help one to learn how to effortlessly move back-and-forth between these internal and external viewponts - a powerful skill that will prove very handy for many related algebraic problems. – Math Gems Jan 24 '13 at 17:12
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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.

Barbara Osofsky
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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 05:27
  • 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
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I think this answer might essentially be the same as @orangeskid's answer, but I'm not sure, so I figured I will post it here. If it is the same or incorrect, please let me know and I will delete it.

Let $I = \langle x_1 - a_1, \dots, x_n - a_n \rangle \subset k[x_1, \dots, x_n]$ for $k$ a field. Consider any ideal $I \subsetneq J$, by the Hilbert Basis theorem $J$ is finitely generated, so we can say that $J=\langle x_1 - a_1, \dots, x_n -a_n, f_1, \dots, f_r\rangle$ for finitely many $1 \le i \le r < \infty$, with $f_i \not\in I, f_i \in k[x_1, \dots, x_n]$ and without loss of generality, all of the $f_i$ are not identically equal to zero.

Case 1: At least one of the $f_i$ is constant.

Without loss of generality, $f_r$ is constant, say $f_r \equiv c \in k, c\not=0$ so $J= \langle x_1 - a_1, \dots, c \rangle = \langle x_1 - a_1, \dots, 1 \rangle = k[x_1, \dots, x_n]$.

Case 2: None of the $f_i$ are constant.

Let's choose one of the $f_i$, which by an abuse of notation we will denote simply by $f_i(x_1, \dots, x_n)$.

Then we have that $$f_i(x_1, \dots, x_n) = f_i^m(x_1, \dots, x_n) + f_i^{m-1}(x_1 -a_1,\dots, x_n - a_n)+ \dots + f_i^1(x_1 -a_1, \dots, x_n -a_n)+ f_i(x_1-a_i,\dots,x_n-a_n) $$ where for all $1 < j \le m, \quad f_i^j(x_1, \dots, x_n) := f_i^{j-1}(x_1,\dots,x_n)-f_i^{j-1}(x_1 - a_1, \dots, x_n-a_n)$, and additionally $f_i^1(x_1, \dots, x_n):=f(x_1,\dots,x_n)-f(x_1-a_1, \dots, x_n-a_n)$, and finally that $m$ is the lowest natural number such that $f_i^m(x_1, \dots, x_n)$ is identically constant, thus for any $k >m$, $f_i^k(x_1, \dots, x_n)\equiv 0$. Such an $m$ is guaranteed to exist, because by construction, $\deg(f_i^1) < \deg(f_i)$ and $\deg(f_i^j) < \deg(f_i^{j-1})$ for all $2 \le j \le m$.

Define $c$ to be the non-zero constant such that $f_i^m(x_1, \dots, x_n) \equiv c $. Now clearly we have that $$ f_i(x_1 -a _1, \dots, x_n -a_n) \in \langle x_1 -a_1, \dots, x_n -a_n \rangle \\ f_i^j(x_1-a_1, \dots, x_n -a_n) \in \langle x_1 - a_1, \dots, x_n - a_n \rangle \quad \forall\ 1\le j \le m-1$$ Therefore we have shown that $f_i(x_1, \dots, x_n) = c + g_i (x_1, \dots, x_n)$ for some $g_i \in \langle x_1 - a_1 , \dots, x_n - a_n \rangle$. Therefore we have that $$ J = \langle x_1 - a_1, \dots, x_n - a_n, \dots, f_i, \dots, f_r \rangle \\ = \langle x_1 - a_1, \dots, x_n - a_n, \dots, c + g_i, \dots, f_r \rangle \\ = \langle x_1 - a_1, \dots, x_n - a_n, \dots, c, \dots, f_r \rangle \\ = \langle x_1 - a_1, \dots, x_n - a_n, \dots, 1, \dots, f_r \rangle = k[x_1, \dots, x_n]$$

Thus, in both cases, $I \subsetneq J \implies J = k[x_1, \dots, x_n]$, so therefore $I$ must be maximal, since clearly $I=\langle x_1 - a_1, \dots, x_n - a_n\rangle$ is proper.

Chill2Macht
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    Since the ring homomorphism $k[x_1,\dots,x_n]\to k[x_1,\dots,x_n]$ such that $x_i\mapsto x_i-a_i$ is an isomorphism, it is not restrictive to assume $a_1=a_2=\dots=a_n=0$. Thus a polynomial $f\notin I$ must have a nonzero constant term and so an ideal properly containing $I$ is the whole ring. This is what your long proof boils down to. And DonAntonio's proof is the abstract (and simpler) version of it. – egreg Oct 11 '16 at 13:04
  • @egreg Thank you for reading my proof. That $x_i \mapsto x_i - a_i$ is an isomorphism (of commutative rings? I'm not sure which type of isomorphism you mean) seems similar to the intuitive idea I had originally involving affine changes of coordinates "not changing anything important". You are probably right that it is equivalent to DonAntonio's proof as well; admittedly I am not comfortable enough with the material to understand the argument why without thinking about it for a long time. I agree that the elegance and simplicity of DonAntonio's proof is preferable. – Chill2Macht Oct 11 '16 at 13:20
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    A ring homomorphism $k[x_1,\dots,x_n]\to A$ is determined as soon as you have a ring homomorphism $k\to A$ and assign images to $x_1,\dots,x_n$. In this case $A=k[x_1,\dots,x_n]$ the homomorphism $k\to A$ is the canonical embedding. With $x_i\mapsto x_i-a_i$ you get an isomorphism, because the inverse is the one doing $x_i\mapsto x_i+a_i$. – egreg Oct 11 '16 at 13:41
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$\textbf{Hint:}$ an ideal $M \subset R$ in a ring is maximal if and only if $R/M$ is a field, can you show if you mod out you get a field?

MyMathYourMath
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