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First of all, I don't think this is the smartest question and I feel quite ashamed about it, but here goes nothing:

I wanted to prove that in the polynomial ring $k[x_1, \ldots, x_n]$ the ideal $I = (x_1 - a_1, \ldots, x_n - a_n)$ is maximal and tried as follows: I have defined a map

$$\varphi: k[x_1, \ldots, x_n] \to k: f(x_1, \ldots, x_n) \mapsto f(a_1, \ldots, a_n)$$ for which I was able to prove that it is a surjective ring homomorphism. Hence I know from the first isomorphism theorem that

$$k[x_1, \ldots, x_n]/ \ker(\varphi) \cong k$$ showing that $\ker(\varphi)$ is maximal. I see that the ideal $I \subset \ker(\varphi)$ but I am stuck on the other direction: suppose $f \in \ker(\varphi)$, then $f(a_1, \ldots, a_n) = 0$. How do I show that it must be of the form $(x_1-a_1)g_1 + \ldots (x_n - a_n)g_n$ for $g_i \in k[x_1, \ldots, x_n]$?

$\textbf{Remark: }$ I know the question on how to prove this is a maximal ideal has been asked many times before, e.g. here and here, but they do not seem to adress my particular question.

Community
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Student
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  • An ideal $I\subset R$ is maximal if and only if $R/I$ is a field. Prove that and you're done. – Justin Young Apr 01 '17 at 11:34
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    Since $k[x_1,x_2,...,x_n] \cong k[x_1-a_1,...,x_n-a_n]$ hence WLOG you can assume $a_i=0 \forall i $. – Math Lover Apr 01 '17 at 11:37
  • @JustinYoung: I know that this holds (I used it to conclude that the kernel has to be maximal), but how does it follow that the ideal $I$ has to be the kernel? – Student Apr 01 '17 at 11:39
  • @MathLover: It took me some time to realize that your comment helps me, thanks! – Student Apr 01 '17 at 11:42
  • I wonder how many times should be this solved on M.SE as the OP's and answerers to stop asking, respectively answer it? – user26857 Apr 01 '17 at 14:17
  • @user26857 I must have missed this question already answering mine: typing the title of my question only showed the links I included in my question. If I had been aware, i would not have asked this egain, really sorry! – Student Apr 01 '17 at 14:52
  • To put the hint above another way, any polynomial can be written as $p(x) = c + (x_1-a_1)q_1 + \cdots + (x_n-a_n)q_n$ for some polynomials $q_i$. This should be clear from the isomorphism stated above. Now you can see that if $p(a) = 0$, then $c=0$. – Justin Young Apr 02 '17 at 15:07
  • @JustinYoung: is this something which we can always do? I mean, is there some result which tells us that in a polynomial ring $k[x_1, \ldots, x_n]$ we can write any polynomial is how you wrote $p(x)$? – Student Apr 02 '17 at 15:21
  • Try to prove it by induction on $n$, it shouldn't be too hard. – Justin Young Apr 02 '17 at 23:04

2 Answers2

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For any polynomial $ f(x_1, x_2, \ldots, x_n) $, you have that

$$ f(x_1, x_2, \ldots, x_n) \equiv f(a_1, a_2, \ldots, a_n) \pmod{I} $$

Thus, $ f \in I $ if and only if the constant polynomial on the RHS is in $ I $, but the only constant polynomial in $ I $ is the zero polynomial. The claim follows.

Ege Erdil
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The congruence relation defined by $I$ is a fairly simple one. It implies $x_i \equiv a_i \bmod I$, and thus

$$ f(\vec{x}) \equiv f(\vec{a}) \pmod I $$

Any two polynomials with the same image under $\varphi$ are equivalent modulo $I$, and so $\ker \varphi \subseteq I$.

Alternatively, $I$ has a simple division algorithm so that it's really easy to pick out a system of reduced representatives for the cosets of $I$: the constant polynomials.

  • Thank you very much! Your answer is very similar to Starfall's answer, which I have accepted. Since I could only accept one answer, I have upvoted yours. – Student Apr 01 '17 at 11:44