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Let $V\subset k^N$ is a finite affine variety (strictly speaking, affine algebraic set), not necessarily irreducible. Show that if $|V|=n$, then $k[V]$, as $k$-algebra, is isomorphic to $k^n$.


I know that $k[V]$ is isomorphic to $k[X]/{\cal{I}} (V)$, writing $V=\{v_1,...v_n\}$, then ${\cal{I}} (V)={\cal{I}} (\{v_1\})\cap {\cal{I}} (\{v_2\})\cap...\cap {\cal{I}} (\{v_n\})$. But I am stuck at showing ${\cal {I}}(V)$ is isomorphic to $k^{N-n}$.

Please help, thank you!

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  • $\begingroup$ You need to assume that $V$ is reduced for this to be correct (that is $I(V)$ is equal to its radical), but maybe this is part of your definition. For example, if $V$ is the affine variety in $\Bbb A^1$ corresponding to the ideal $(x^2)$, we have $|V| = 1$ and $k(V) \cong k^2$. $\endgroup$ Commented Dec 10, 2017 at 15:05
  • $\begingroup$ @NicolasHemelsoet Could you explain why $k(V) \cong k^2$? $\endgroup$
    – user771160
    Commented Dec 11, 2017 at 4:40
  • $\begingroup$ Sure, we have $k[V] \cong k[x]/(x^2) \cong k \oplus k{x} \cong k^2$. $\endgroup$ Commented Dec 11, 2017 at 9:47

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Define the following homomorphism: $f:k[X]\rightarrow k^n,\quad p\mapsto (p(v_1),\ldots,p(v_n))$. Show that $f$ has kernel $\mathcal{I(V)}$ and is surjective and apply the fundamental theorem on homomorphisms.

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  • $\begingroup$ That is exactly what I want. Thanks! $\endgroup$
    – user771160
    Commented Dec 10, 2017 at 14:25

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