Lemma 33.34.2. Let $k$ be a field and $n \geq 0$. Let $X, Y \subset \mathbf{A}^ n_ k$ be closed subsets. Assume that $X$ and $Y$ are equidimensional, $\dim (X) = r$ and $\dim (Y) = s$. Then every irreducible component of $X \cap Y$ has dimension $\geq r + s - n$.
Proof. Consider the closed subscheme $X \times Y \subset \mathbf{A}^{2n}_ k$ where we use coordinates $x_1, \ldots , x_ n, y_1, \ldots , y_ n$. Then $X \cap Y = X \times Y \cap V(x_1 - y_1, \ldots , x_ n - y_ n)$. Let $t \in X \cap Y \subset X \times Y$ be a closed point. By Lemma 33.20.5 we have $\dim _ t(X \times Y) = \dim (X) + \dim (Y)$. Thus $\dim (\mathcal{O}_{X \times Y, t}) = r + s$ by Lemma 33.20.3. By Algebra, Lemma 10.60.13 we conclude that
\[ \dim (\mathcal{O}_{X \cap Y, t}) = \dim (\mathcal{O}_{X \times Y, t}/(x_1 - y_1, \ldots , x_ n - y_ n)) \geq r + s - n \]
This implies the result by Lemma 33.20.3. $\square$
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