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Let $V$ be an irreducible variety of dimension 0 over an algebraically closed field $k$.

I understand that the function field $K(V)$ is equal to $k$. The answer to algebraic variety of dimension 0 uses this fact to conclude that $k[x_1,\dots,x_n]/I_V=k$, where $I_V$ is the ideal of $V$. Why is this true?

(The definitions I am using are the same as those in the linked question.)

rpf
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    It is always the case that $K(V) = \text{Frac}(K[V])$ where $K[V] = K[x_1,\ldots,x_n]/I_V$, $I_V= { f \in K[x_1,\ldots,x_n], \forall a \in V, f(a) = 0}$. Here $K[V] \cong K$ thus $K(V) = K[V]$ and $I_V= (x_1-a_1,\ldots,x_n-a_n)$. – reuns Oct 07 '17 at 01:14
  • I thought that $K(V)=\text{Frac}(K[V])$ only holds when $V$ is an affine variety. If this is true for all varieties, where may I find a proof? – rpf Oct 07 '17 at 01:22
  • Sorry I meant irreducible affine variety. For a projective variety you need an homogeneous prime ideal, the fraction field of the ring it generates, and the subfield of projective rational functions. – reuns Oct 07 '17 at 01:29

2 Answers2

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Since $V$ is irreducible, the algebra of functions $k[x_1, \ldots, x_n]/I_V$ embeds into the function field $K(V) = k$. This means that we have an injective map of $k$-algebras $$ k[x_1, \ldots, x_n]/I_V \to k$$ which has image $k$, and so by the isomorphism theorem $k[x_1, \ldots, x_n]/I_V \cong k$.

Joppy
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  • $V$ is irreducible thus it has a function field $K(V)$ – reuns Oct 07 '17 at 01:25
  • Why is the injection $k[x_1,\dots,x_n]/I_Z\to k$ surjective? – rpf Oct 07 '17 at 01:47
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    It is a map of $k$-algebras, and so $k$ is always preserved. Alternatively, the only ideal of the polynomial ring containing $k$ is the unit ideal. – Joppy Oct 07 '17 at 01:57
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Just an addemdum, correct me if I am wrong:

Basically, dimension of $X$ is supremum of lenght $n$ of chains of irreducible subvarieties $X_0\supsetneq \cdots \supsetneq X_n \neq \emptyset $, where $X_n$ is a point (ie, is just possible to construct subvarieties by decreasing dimension). So, dimension is $0$ $\iff$ $X$ is a single point.

Quiet_waters
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