Proposition

A variety $X$ is irreducible if $I(X)$ is prime.

Solution

Let a variety $X$ be given. Suppose that $I(X)$ is a prime ideal. Let $X_1, X_2$ be two varieties such that $X = X_1 \cup X_2$. If $X_1 = X$ or $X_2 = X$, we are done. Suppose otherwise. Let $I_1 = I(X_1)$ and $I_2 = I(X_2)$.

• $X_1 = \{ p \in k^n \mid \forall f \in A, f(p) = 0 \}$. If every $f \in I_1$ vanishes on $X$, then $X \subset X_1$. Since we assume that $X \ne X_1$, there exists $f_1 \in I_1$ such that $f_1$ does not vanish on $X$. $f_1$ obviously vanishes on $X_1$ because $f_1 \in I_1$.
• Similarly, there exists $f_2 \in I_2$ such that $f_2$ does not vanish on $X$. $f_2$ obviously vanishes on $X_2$ because $f_2 \in I_2$.

Consider $f = f_1f_2$. Let $p \in X$. If $p \in X_1$, then $f(p) = f_1(p)f_2(p) = 0 \cdot f_2(p) = 0$. If $p \in X_2$, then $f(p) = f_1(p)f_2(p) = f_1(p) \cdot 0 = 0$. Thus $f$ vanishes on $X$. $f \in I(X)$, but $f_1 \notin I(X)$ and $f_2 \notin I(X)$ because they do not vanish on $X$. This is a contradiction because $I(X)$ is a prime ideal. Therefore, $X$ must be irreducible.