Let $k$ be an algebraically closed field, And let $V(\mathfrak{a})$ be the algebraic variety generated by the ideal $\mathfrak{a} \subset k[x_1,\cdots,x_n]$. I have read that we can identify $V(\mathfrak{a})$ as

$V(\mathfrak{a})=\text{Hom}_{k-alg}( k[x_1,\cdots,x_n]/ \mathfrak{a} , k)$.

But I don't see how. I know that the points of $V(\mathfrak{a})$ is in bijection with the maximal ideals of $k[x_1,\cdots,x_n]/ \mathfrak{a}$ by Hilbert's Nullstellenstaz, but I don't see how to view it as $k$-algebra homomorphism.

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    $\begingroup$ The kernal of a morphisn $k[x]/a \to k$ is a maximal ideal. $\endgroup$ – Youngsu Dec 7 '18 at 9:06
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    $\begingroup$ Adding to the above comment: An algebra homomorphism from a polynomial algebra is uniquely determined by where it sends the variables. Taking a quotient then introduces a suitable requirement that certain combinations of these values are 0. $\endgroup$ – Tobias Kildetoft Dec 7 '18 at 9:35
  • $\begingroup$ Hi can you explain more I'm still a bit confused.. $\endgroup$ – Ishigami Dec 7 '18 at 14:19
  • $\begingroup$ I remember you that the homomorphism of $\mathbb{K}$-algebras send the units in the units; so the homomorphism from $\mathbb{K}[x_1,\dots,x_n]/\mathfrak{a}$ to $\mathbb{K}$ are all "onto". From all this, using the comment of @Youngsu, you have the statement; for exact, $V(\mathfrak{a})$ is in bijection with the set $Hom_{\mathbb{K}-alg}(\mathbb{K}[x_1,\dots,x_n]/\mathfrak{a},\mathbb{K})$. $\endgroup$ – Armando j18eos Dec 13 '18 at 13:47

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