Affine Varieties – The Math


Definition: Let $K$ be a field and let $X subseteq mathbb{A}^n(K)$ be an affine algebraic set. Then $X$ is said to be Reducible if there exists affine algebraic sets $X_1$ and $X_2$ where $X_1, X_2 neq emptyset$ and $X_1, X_2 neq X$ and such that $X = X_1 cup X_2$. An affine algebraic set $X$ is said to be Irreducible if it is not reducible.

For example consider the affine $2$-space $mathbb{A}^2(mathbb{R})$ and let $X = V(x(y-x))$. Then $x(y-x) = 0$ if and only if $x = 0$ or $y = x$. We can graph this affine algebraic set in $mathbb{R}^2$:

Screen%20Shot%202017-01-29%20at%203.55.24%20AM.png

It can easily be verified that if $X_1 = V(x)$ and $X_2 = V(y – x)$ then:

(1)

begin{align} quad X = X_1 cup X_2 end{align}

So $X$ is a reducible affine algebraic set.

We are about to prove an important result which will allow us to determine whether or not a affine algebraic set is irreducible or not. We first need to make a definition.

Definition: Let $R$ be a ring and let $I$ be an ideal. Then $I$ is said to be a Prime Ideal if whenever $ab in I$ we have that either $a in I$ or $b in I$.
Theorem 1: Let $K$ be a field and let $X subseteq mathbb{A}^n(K)$ be an affine algebraic set. Then $X$ is irreducible if and only if $I(X)$ is a prime ideal.
  • Proof: $Rightarrow$ Let $X$ be an irreducible affine algebraic set and suppose that $I(X)$ is not a prime ideal. Then there exists elements $F, G not in I(X)$ such that $FG in I(X)$. Now since $FG in I(X)$ we have that $(FG) subseteq I(X)$. Therefore:

(2)

begin{align} quad V((FG)) supseteq V(I(X)) end{align}

  • But $V((FG)) = V(F) cup V(G)$. And since $X$ is an affine algebraic set we have that $X = V(I(X))$. So from above we see that:

(3)

begin{align} quad V(F) cup V(G) supseteq X end{align}

  • Let $X_1 = V(F) cap X$ and let $X_2 = V(G) cap X$. Then $X_1$ and $X_2$ are affine algebraic sets. Furthermore, $X = X_1 cup X_2$. Also, $X_1, X_2 neq emptyset$ and $X_1, X_2 neq X$. So $X$ is a reducible affine algebraic set which is a contradiction. So the assumption that $I(X)$ is not a prime ideal is false. So $I(X)$ is a prime ideal.
  • $Leftarrow$ Let $I(X)$ be a prime ideal and suppose that $X$ is a reducible affine algebraic set. Then there exists affine algebraic sets $X_1, X_2$ such that $X_1, X_2 neq emptyset$ and $X_1, X_2 neq X$ with:

(4)

begin{align} quad X = X_1 cup X_2 end{align}

  • So $X_1 subset X$ and $X_2 subset X$. Therefore $I(X_1) supset I(X)$ and $I(X_2) supset I(X)$. So there exists functions $F in I(X_1) setminus I(X)$ and $G in I(X_2) setminus I(X)$. Consider the function $FG$. Then $FG$ vanishes on $X_1$ and on $X_2$. So $FG$ vanishes on $X_1 cup X_2$. So $FG$ vanishes on $X$, i.e., $FG in I(X)$. But $F, G not in I(X)$. This contradicts $I(X)$ being a prime ideal. So the assumption that $X$ is reducible was false. Hence $X$ is an irreducible affine algebraic set. $blacksquare$

The affine algebraic sets which are irreducible will be given a special name.

Definition: Let $K$ be a field. An Affine Variety if an irreducible affine algebraic set.

From the example above, we see that if $X = V(x(y-x)) subseteq mathbb{A}^2(mathbb{R})$ then $X$ is not an affine variety.



The math online

Leave a Comment

Your email address will not be published. Required fields are marked *