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I am trying to answer the question already asked here.

My question is two parts:

1) I think I have found a proof on my own, could someone check it is valid?

Modulo that ideal, $x_i\equiv a_i$ so any polynomial in $k[x_1,\cdots, x_n]$ is congruent to the polynomial evaluated at $(a_1,\cdots,a_n)$ which is just an element of $k.$ Hence non-zero elements are invertible, so the quotient is a field and the ideal is maximal.

2) Trying to following DonAntonio's answer, I can't actually compute the kernel of that homomorphism. I can show that ideal is in the kernel, but not the reverse inclusion. Any hints?

Thanks!

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  • $\begingroup$ @GitGud No worries. $\endgroup$ – Katie Dobbs Mar 10 '13 at 11:35
  • $\begingroup$ Yes, that's essentially the hint I gave to the prior questions (see also the comments to my answer). $\endgroup$ – Math Gems Mar 11 '13 at 3:19
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1) Yes, correct thinking.

2) So, $f:k[x_1,..,x_n]\to k \ \ (g(x_1,..,x_n)\mapsto g(a_1,..,a_n))$ , and assume that a polynomial $g$ is in its kernel, i.e. $g(a_1,..,a_n)=0$. Probably the easiest if we reduce it to the case where all $a_i=0$. For that end, set $z_i:=x_i-a_i$ and replace every $x_i$ by $z_i+a_i$ in each monomial of $g$, and expand them. Thus we have another polynomial $h(z_1,..,z_n)=g(x_1,..,x_n)$ with the property that $h(0,0,..,0)=0$, i.e. the constant term of $h$ is $0$. And it is now readily seen that this implies $h$ is in the ideal $(z_1,..,z_n)$ of $k[z_1,..,z_n]$.

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  • $\begingroup$ Ahh yes. Thank you very much! $\endgroup$ – Katie Dobbs Mar 10 '13 at 11:41

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