# Viewing algebraic varieties as Homomorphism

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.

• The kernal of a morphisn $k[x]/a \to k$ is a maximal ideal. – Youngsu Dec 7 '18 at 9:06
• 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. – Tobias Kildetoft Dec 7 '18 at 9:35
• Hi can you explain more I'm still a bit confused.. – Ishigami Dec 7 '18 at 14:19
• 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 $Hom_{\mathbb{K}-alg}(\mathbb{K}[x_1,\dots,x_n]/\mathfrak{a},\mathbb{K})$. – Armando j18eos Dec 13 '18 at 13:47