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This question already has an answer here:

$K$ is field. $a_1$,$a_2$ elements of $K$. Show that $(x_1-a_1,x_2-a_2)$ is a maximal ideal of $K[x_1,x_2]$.

$K[x_1,x_2]$ is UFD so if $K[x_1,x_2]/(x_1-a_1,x_2-a_2)$ is field then $(x_1-a_1,x_2-a_2)$ is maximal ideal.

If I can show that $K[x_1,x_2]/(x_1-a_1,x_2-a_2)$ isomorphic to $K$, we can verify that $(x_1-a_1,x_2-a_2)$ maximal ideal of $K[x_1,x_2]$.

thanks for helps and comments.

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marked as duplicate by user26857, Ken Duna, Arnaud D., C. Falcon, Namaste abstract-algebra Apr 2 '17 at 0:22

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    $\begingroup$ Hint: $$\phi:K[x_1,x_2]/\langle x_1-a,x_2-b\rangle \to K\;,\;\phi(f(x,y):=f(a,b)\;\ldots$$ $\endgroup$ – DonAntonio Nov 3 '13 at 22:06
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    $\begingroup$ $K[x,y]/(x-a_1, y-a_2)$ is just $K[x,y]$ modulo the relations $x=a_1$ and $y=a_2$. Of course it is isomorphic to $K$. You can fill in the details. $\endgroup$ – TBrendle Nov 3 '13 at 22:08
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Consider the following map (slightly modified the suggestion of @DonAntonio in comments): $$\phi:K[x,y]\to K,\quad \phi(f(x,y)):=f(a,b)$$ Then show that its kernel is just the ideal $(x-a,\,y-b)$ and its image contains $1$, so it is surjective, then use the first isomorphism theorem.

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