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I am trying to solve the following problem.

If $ \mathcal{K}$ is a field and $a_1,a_2,\dots,a_n \in \mathcal{K}$. Prove that $(x_1-a_1,x_2-a_2,\dots,x_n-a_n)$ is a maximal ideal in $\mathcal{K}[x_1-a_1,x_2-a_2,\dots,x_n-a_n]$.

I know that if $R$ is a ring and $I$ and ideal in $R$, $R/I$ is a field if and only if $I$ is a maximal ideal.

So I need a surjective homomorphism $\phi \colon \mathcal{K}[x_1-a_1,x_2-a_2,\dots,x_n-a_n] \to \mathcal{K}$ such that $\text{ker}\phi =(x_1-a_1,x_2-a_2,\dots,x_n-a_n) $ so I can use the first theorem of homomorphism of rings and use the former resulst and I am done!

I think the evaluation is the homomorphism that I am looking for but I cant prove that indeed is a homomorphism... Can you help me with that please.

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  • $\begingroup$ You simply want to show that evaluation of polynomials at a point gives us a ring homomorphism? $\endgroup$ – hardmath May 28 '15 at 20:10
  • $\begingroup$ that is also surjetive $\endgroup$ – KaushL May 28 '15 at 20:19
  • $\begingroup$ Surjectivity is easy -- what does evaluation do to the constants of the polynomial ring? $\endgroup$ – hardmath May 28 '15 at 20:20
  • $\begingroup$ What if start with $a_i=0$ for all $i$? They complicate the question in an unnecessary way. $\endgroup$ – user26857 May 28 '15 at 20:24
  • $\begingroup$ I dont get it.. $\endgroup$ – KaushL May 28 '15 at 20:28
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The question asks from you to observe first that there is an isomorphism between $K[X_1,\dots,X_n]$ and $K[X_1-a_1,\dots,X_n-a_n]$ sending $X_i$ to $X_i-a_i$. Under this isomorphism the ideal $(X_1,\dots,X_n)$ corresponds to the ideal $(X_1-a_1,\dots,X_n-a_n)$, so $$K[X_1,\dots,X_n]/(X_1,\dots,X_n)\simeq K[X_1-a_1,\dots,X_n-a_n]/(X_1-a_1,\dots,X_n-a_n).$$ This reduces the problem to the following:

Prove that $(X_1,\dots,X_n)$ is a maximal ideal in $K[X_1,\dots,X_n]$.

The surjective homomorphism you need now is $K[X_1,\dots,X_n]\to K$ defined by $X_i\mapsto 0$ for all $i$.

Edit. I've noticed now that you have troubles with proving that the evaluation is indeed a homomorphism. It seems you have to learn about the universal property of polynomial rings: google it or look here.

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The evaluation is indeed the map that you are looking for. To show that it is a homomorphism is straightforward (especially additivity). Take two polynomials, add and multiply them, and then evaluate. Also $$(x_1-a_1,\dots ,x_n-a_n)\subseteq \text{Ker}(\phi)$$ should be obvious. The only kind of tricky part of this exercise is the other inclusion.

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