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I am reading Undergraduate algebraic geometry by Miles Reid and over there I can see two definitions of affine variety over algebraically closed field $k$. I want to know how these two definitions are equivalent,

(a)An affine variety is an algebraically irreducible subset, defined upto isomorphism.

(b) An affine variety over a field $k$ is a set $V$, together with a ring $k[V]$ of k-valued functions $f: V —> k$ such that

(i) $k[V]$ is a finitely generated k-algebra,

(ii)for some choice of genertators $x_1,x_2,....x_n$ of $k[v]$ over $k$, the map $f: V —> A_n^k$ by

$f(p)=x_1(p),x_2(p),....x_n(p)$

embeds V as an irreducible algebraic set. I want to know how these two definitions are equivalent, namely if every algebraically irreducible set can be embedded in this way. Also, does by k-valued function means they are all polynomial functions. It will be great if you can help me with this doubt!

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  • $\begingroup$ I think for (b), you also need that $k[V]$ is a reduced $k$-algebra as well. $\endgroup$
    – daruma
    Jun 6, 2021 at 2:17
  • $\begingroup$ Nothing of this sort is written. But I can't understand how can we say every algebraically closed set can be embedded this way? I thought I can do it by projective functions, but ring of k-valued functions may include non polynomial functions also $\endgroup$
    – user631697
    Jun 6, 2021 at 2:40
  • $\begingroup$ I am not entirely sure what you had in mind when you said nonpolynomial functions. If you want to look at functions defined only on some open subset of an affine algebraic set, you can allow rational functions. But if you want functions defined globally on all of $V$, you have to have polynomials. $\endgroup$
    – daruma
    Jun 6, 2021 at 2:56
  • $\begingroup$ Since $V$ is embedded algebraically by generators of $k[V]$, those generators satisfy polynomial relations, showing that $k[V]$ is a polynomial ring; this is the sense in which the elements of $k[V]$ "are" polynomials. $\endgroup$ Jun 6, 2021 at 3:07
  • $\begingroup$ I meant that in K[V] there may be polynomial functions which isn't generated by X_is where X_i is a function projecting i-th coordinate $\endgroup$
    – user631697
    Jun 6, 2021 at 3:16

1 Answer 1

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I don't have Miles Reid's UAG with me but having read the Commutative Algebra book by him this is how he defines things.

When one defines an affine algebraic set, the first way that it's usually defined is the zero set of an ideal $I$ of a polynomial ring. With this definition one defines an algebraic set as a subset of an affine $n$-space.

i.e. $I\subset k[X_1,..., X_n]$

Irreducibility tells us that $I$ is a prime ideal. This follows from the relationship of unions/intersections of ideals and algebraic sets.

When we have an affine algebraic set or an affine variety, we want to look at what the functions are on it. Since we are doing algebraic geometry, really we care about algebraic functions, that is to say we are interested in polynomials.

However, when we talk about functions, we don't usually make a distinction when two functions take the same value everywhere on your affine algebraic set or your affine variety. For example, the function $y-x^2$ is a polynomial but its value is $0$ everywhere on $V(y-x^2)$. So really we want to identify this as the same as the zero function. That is why when we look at the affine coordinate ring of $V$, we define it as $k[V]:=k[X_1,...,X_n]/I$ (where $I=I(V)$) so that we identify the functions whose values agree everywhere on $V$.

Now what can we say when we look at $k[V]$ from the perspective of commutative algebra? Firstly, $k[V]$ is a quotient of a polynomial ring in $n$ indeterminates so we can say that it is finitely generated by the images of the indeterminates in the quotient. (Warning this means, finitely generated as a $k$-algebra not a $k$-module).

In the case of affine varieties, we also have that $I=I(V)$ is a prime ideal. When we quotient out by a prime ideal, we don't have any zero divisors in $k[X_1,...,X_n]/I$.

So the first definition naturally gives you the second definition because you already choose how you embed $V$ into $\mathbb{A}^n$ and your ring of functions on $V$ can be interpreted very concretely as the affine coordinate ring.

The second definition is in some sense more natural, as we don't choose an explicit embedding of $V$. For example, you can embed $\mathbb{A}^n\subset \mathbb{A}^{m}$ for any $m\geq n$ via $(a_1,...,a_n)\mapsto (a_1,...,a_n,0,..,0)$. We face the same issue for other affine algebraic varieties and affine algebraic sets. But you can see that there are other ways of embedding this $n$-space into $m$-space. The problem is that there is no canonical way of embedding your affine algebraic variety into an affine space.

So the second definition regards the variety as an abstract set with a ring of functions $k[V]$. (ii) of the second definition is saying there is an embedding of $V$ into an affine space but note that we are not saying this is the correct or canonical choice of embedding.

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  • $\begingroup$ Sorry, small error in the previous version. I said reduced rings were rings with no zero divisors when it should say nilpotents. $\endgroup$
    – daruma
    Jun 10, 2021 at 0:31

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