Affine Varieties

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 $\mathbb{R}^2$:

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.