[proofplan] We prove both directions of the equivalence. For the forward direction, we take $fg \in I(X)$ and use the fact that $X \subseteq V(f) \cup V(g)$ to decompose $X$ as a union of two algebraic subsets; irreducibility forces $X$ into one of them, giving $f \in I(X)$ or $g \in I(X)$. For the converse, we take a decomposition $X = X_1 \cup X_2$ into algebraic sets, use $I(X) = I(X_1) \cap I(X_2)$, and show that a prime ideal equal to an intersection of two ideals must equal one of them, so the decomposition is trivial. [/proofplan] [step:Show that irreducibility of $X$ implies $I(X)$ is prime] Assume $X$ is irreducible. Let $f, g \in \Omega[T_1, \dots, T_n]$ with $fg \in I(X)$. We must show $f \in I(X)$ or $g \in I(X)$. For every $\underline{x} \in X$, we have $(fg)(\underline{x}) = f(\underline{x}) g(\underline{x}) = 0$, so $f(\underline{x}) = 0$ or $g(\underline{x}) = 0$. Therefore \begin{align*} X \subseteq V(f) \cup V(g). \end{align*} Writing $X = (X \cap V(f)) \cup (X \cap V(g))$, both $X \cap V(f)$ and $X \cap V(g)$ are $\Omega$-algebraic subsets of $X$: each is the intersection of $X$ with a vanishing set, and the intersection of two algebraic sets is algebraic. Since $X$ is irreducible, this decomposition forces $X = X \cap V(f)$ or $X = X \cap V(g)$, i.e., $X \subseteq V(f)$ or $X \subseteq V(g)$. If $X \subseteq V(f)$, then $f(\underline{x}) = 0$ for all $\underline{x} \in X$, so $f \in I(X)$. Similarly, if $X \subseteq V(g)$, then $g \in I(X)$. Therefore $I(X)$ is a prime ideal. [guided] Assume $X$ is irreducible. Recall that an $\Omega$-algebraic set is irreducible if it cannot be written as a union $X = X_1 \cup X_2$ of two proper algebraic subsets. We want to show $I(X)$ is prime, i.e., whenever $fg \in I(X)$, either $f \in I(X)$ or $g \in I(X)$. Take $f, g \in \Omega[T_1, \dots, T_n]$ with $fg \in I(X)$. The condition $fg \in I(X)$ means that for every $\underline{x} \in X$, we have $f(\underline{x})g(\underline{x}) = 0$. Since $\Omega$ is a field (hence an integral domain), for each $\underline{x}$ either $f(\underline{x}) = 0$ or $g(\underline{x}) = 0$. This means every point of $X$ lies in $V(f)$ or $V(g)$, i.e., \begin{align*} X \subseteq V(f) \cup V(g). \end{align*} Intersecting both sides with $X$, we can write $X = (X \cap V(f)) \cup (X \cap V(g))$. Now $X \cap V(f)$ and $X \cap V(g)$ are algebraic sets: if $X = V(J)$ for some ideal $J$, then $X \cap V(f) = V(J + (f))$, which is an algebraic set. Similarly for $X \cap V(g)$. Since $X$ is irreducible, a decomposition of $X$ as a union of two algebraic subsets forces one of them to equal $X$. So $X = X \cap V(f)$ or $X = X \cap V(g)$, which means $X \subseteq V(f)$ or $X \subseteq V(g)$. If $X \subseteq V(f)$, then $f$ vanishes at every point of $X$, so $f \in I(X)$. If $X \subseteq V(g)$, then $g \in I(X)$. Either way, $f \in I(X)$ or $g \in I(X)$, which is exactly the statement that $I(X)$ is prime. [/guided] [/step] [step:Show that $I(X)$ prime implies $X$ is irreducible via the identity $I(X_1 \cup X_2) = I(X_1) \cap I(X_2)$] Assume $I(X)$ is a prime ideal. Suppose $X = X_1 \cup X_2$ where $X_1$ and $X_2$ are $\Omega$-algebraic subsets of $X$. We must show $X_1 = X$ or $X_2 = X$. Since $I(\cdot)$ reverses inclusions and converts unions to intersections for algebraic sets, we have \begin{align*} I(X) = I(X_1 \cup X_2) = I(X_1) \cap I(X_2). \end{align*} [claim:A prime ideal equal to an intersection of two ideals equals one of them] Let $R$ be a commutative ring and $\mathfrak{p}$ a prime ideal with $\mathfrak{p} = \mathfrak{a}_1 \cap \mathfrak{a}_2$ for ideals $\mathfrak{a}_1, \mathfrak{a}_2 \trianglelefteq R$. Then $\mathfrak{p} = \mathfrak{a}_1$ or $\mathfrak{p} = \mathfrak{a}_2$. [/claim] [proof] Suppose $\mathfrak{p} \neq \mathfrak{a}_1$ and $\mathfrak{p} \neq \mathfrak{a}_2$. Then there exist $a_1 \in \mathfrak{a}_1 \setminus \mathfrak{p}$ and $a_2 \in \mathfrak{a}_2 \setminus \mathfrak{p}$. The product $a_1 a_2 \in \mathfrak{a}_1 \cap \mathfrak{a}_2 = \mathfrak{p}$. Since $\mathfrak{p}$ is prime and $a_1 a_2 \in \mathfrak{p}$, either $a_1 \in \mathfrak{p}$ or $a_2 \in \mathfrak{p}$, contradicting the choice of $a_1$ and $a_2$. Therefore $\mathfrak{p} = \mathfrak{a}_1$ or $\mathfrak{p} = \mathfrak{a}_2$. [/proof] Applying the claim to $\mathfrak{p} = I(X)$, $\mathfrak{a}_1 = I(X_1)$, and $\mathfrak{a}_2 = I(X_2)$, we obtain $I(X) = I(X_1)$ or $I(X) = I(X_2)$. If $I(X) = I(X_1)$, then applying $V(\cdot)$ and using the fact that $V(I(Y)) = Y$ for any $\Omega$-algebraic set $Y$ (which holds because $\Omega$ is algebraically closed, by the strong Nullstellensatz applied to ideals of the form $I(Y)$ which are already radical), we get $X = V(I(X)) = V(I(X_1)) = X_1$. Similarly, if $I(X) = I(X_2)$, then $X = X_2$. In either case, the decomposition $X = X_1 \cup X_2$ is trivial (one of $X_1, X_2$ equals $X$). Therefore $X$ is irreducible. [guided] Now assume $I(X)$ is prime, and suppose $X = X_1 \cup X_2$ is a decomposition of $X$ into two $\Omega$-algebraic subsets. We want to show this decomposition is trivial. The vanishing ideal functor $I(\cdot)$ converts unions to intersections: a polynomial vanishes on $X_1 \cup X_2$ if and only if it vanishes on both $X_1$ and $X_2$, so \begin{align*} I(X) = I(X_1 \cup X_2) = I(X_1) \cap I(X_2). \end{align*} So we have a prime ideal $I(X)$ expressed as an intersection of two ideals $I(X_1)$ and $I(X_2)$. Can a prime ideal be a proper intersection? We claim not. Suppose $I(X) \neq I(X_1)$ and $I(X) \neq I(X_2)$. Pick $a_1 \in I(X_1) \setminus I(X)$ and $a_2 \in I(X_2) \setminus I(X)$. Then $a_1 a_2 \in I(X_1) \cap I(X_2) = I(X)$. But $I(X)$ is prime, so $a_1 \in I(X)$ or $a_2 \in I(X)$. Both possibilities contradict our choices. Therefore $I(X) = I(X_1)$ or $I(X) = I(X_2)$. Now we recover the geometric conclusion. If $I(X) = I(X_1)$, apply $V(\cdot)$ to both sides. For an $\Omega$-algebraic set $Y$, we have $V(I(Y)) = Y$: the inclusion $Y \subseteq V(I(Y))$ holds by definition, and $V(I(Y)) \subseteq Y$ follows because $I(Y)$ is a radical ideal (by the strong Nullstellensatz, $I(V(J)) = \sqrt{J}$, and applying this with $J = I(Y)$ which satisfies $\sqrt{I(Y)} = I(Y)$ since $I(Y)$ is always radical, gives $V(I(Y)) = V(\sqrt{I(Y)}) = V(I(Y))$, and $Y = V(I(Y))$ since $Y$ is algebraic). So $X = V(I(X)) = V(I(X_1)) = X_1$. The argument is symmetric: if $I(X) = I(X_2)$, then $X = X_2$. In both cases, the decomposition $X = X_1 \cup X_2$ has one factor equal to $X$, so $X$ is irreducible. [/guided] [/step]