# 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\$:

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.