How do I show that the coordinate ring $K[V]$ is isomorphic to $K[p_1\mid V,\ldots,p_n\mid V]$

Question

Let $K$ be an algebraically closed field.

Let $I:PK^n \rightarrow PK[x]$ be the standard map from subsets to ideals. ($P$ here is the power-set functor - so that is $PK^n$ is all subsets of $K^n$ and $PK[x]$ is all subsets of $K[x]$ ,and where we are writing $x=x_1,\ldots,x_n$. Explicitly $I(V):=\{f\in K[x]: f(V)=0\}$).

Let $V$ be an algebraically closed set in $K^n$.

Then the coordinate ring $K[V]$ is defined to be $K[x]/I(V)$.

Now define $p_i:K^n\rightarrow K$ by $a\rightarrow a_i$ the $i$th projection, and write $p:=(p_1,\ldots,p_n)$

a. How do we show that in fact, $K[V]=K[p\mid V]$? (where $p\mid V$ is the restriction of $p$ to $V$).

This is asserted in Milnes Algebraic Geometry, pg. 47 in the section The coordinate ring of an algebraic set. But I can't see why it is true.

It's quite unclear what you are asking, can you please edit your question? — , Jul 04 '13 at 03:48
Yeah that power-set functor thing is not obvious to me. I think it's better if you make the map explicit. — Patrick Da Silva, Jul 04 '13 at 03:50
@DaSilva:I've edited to make the power-set functor action clearer. — Mozibur Ullah, Jul 04 '13 at 03:54
@BenjaLim: sure, can you point out where you find it confusing? Is it, say confusing that I'm abbreviating $x_1,...,x_n$ by $x$? — Mozibur Ullah, Jul 04 '13 at 03:56

score 1 · Answer 1 · edited Apr 13 '17 at 12:21

I'll piggyback on this recent answer, mostly for the polynomial $\leftrightarrow$ function setup.

The elements $x_i$ generate $K[x]$ over $K$ and hence the quotient $K[x]/I(V)$ over $K$. We're identifying [that this is being done, perhaps implicitly, might be what's tripping you up] this quotient with a subring of $\operatorname{Fun}(V, K)$, so you ask what the images of the $x_i$ are in that ring: what function $V \to K$ does $x_i$ define?

How do I show that the coordinate ring $K[V]$ is isomorphic to $K[p_1\mid V,\ldots,p_n\mid V]$

1 Answers1