# Proving $(x_1 - a_1, \ldots, x_n -a_n)$ is a maximal ideal [duplicate]

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.

## marked as duplicate by user26857 abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 1 '17 at 14:16

• 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 1 '17 at 11:34
• 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 1 '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 1 '17 at 11:39
• @MathLover: It took me some time to realize that your comment helps me, thanks! – Student Apr 1 '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 1 '17 at 14:17

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.

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 1 '17 at 11:44