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I cannot solve the following problem, from Reid Undergraduate Commutative Algebra, exercise 1.16:

Let $(a_1,...,a_n) \in K^n$ where $k \subset K$ an algebraic field extension. Determine the image and kernel of the evaluation map $e_a: k[x_1,...,x_n] \rightarrow K$. Prove that $(x-a_1,...,x-a_n) \cap k[x_1,...,x_n]$ is a maximal ideal of $k[x_1,...,x_n]$.

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  • $\begingroup$ Determine the image and kernel? Have you determined (i) the image, (ii) the kernel? Do you need a hint for either? $\endgroup$ – Lord Shark the Unknown Oct 29 '17 at 12:17
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By definition, the image is $k[a_1, \dotsc, a_n]$. Since all $a_i$ are algebraic over $k$, this is actually a field, i.e the image is $k(a_1, \dotsc, a_n)$.

To compute the kernel, note that the map factors as $$k[x_1, \dotsc, x_n] \hookrightarrow K[x_1, \dotsc, x_n] \xrightarrow{x_i \mapsto a_i} K.$$

The kernel of the second map is well known to be $(x_1-a_1, \dotsc, x_n-a_n)$, hence the kernel of the composition is precisely the ideal mentioned in the second task of your question. It immediately follows that this ideal is maximal, since the image is a field.

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  • $\begingroup$ A concisely, beautifully presented answer: +1. $\endgroup$ – Georges Elencwajg Oct 29 '17 at 12:36

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