Let $X,Y \subset \mathbb{C}^n$ be two irreducible affine algebraic sets with nonempty intersection. I am looking for a reference where I can find the proof that $$\dim (X \cap Y) \geq \dim X+ \dim Y - n. $$ Any suggestion or a proof would be appreciated!
1 Answers
Here's the proof from Hartshorne (modulo a few minor editorial corrections):
Proposition I.7.1 (Affine dimension theorem): Let $Y,Z$ be varieties (respectively irreducible closed subschemes) of dimensions $r,s$ in $\Bbb A^n_k$ for $k$ an algebraically closed field (resp. any field). Then every irreducible component $W$ of $Y\cap Z$ has dimension $\geq r+s-n$.
Proof. We proceed in several steps. First, suppose that $Z$ is a hypersurface given by $V(f)$. If $Y\subset Z$, there is nothing to prove. If $Y\not\subset Z$, then we must show that each irreducible component of $Y\cap Z$ has dimension $r-1$. Let $A(Y)$ be the affine coordinate ring of $Y$. Then the irreducible components of $Y\cap Z$ correspond to the minimal primes $\mathfrak{p}$ lying over the ideal $(f)\subset A(Y)$. By Krull's principal ideal theorem, each such prime has height one, so by the dimension theorem, $A(Y)/\mathfrak{p}$ has dimension $r-1$. This shows that every irreducible component $W$ has dimension $r-1$.
Now for the general case. Consider the product $Y\times Z\subset \Bbb A^{2n}$, which is a variety of dimension $r+s$. Let $\Delta$ be the diagonal $\{P\times P \mid P\in\Bbb A^n\}$. Then $\Bbb A^n$ is isomorphic to $\Delta$ by the map $P\mapsto P\times P$, and under this isomorphism $Y\cap Z$ corresponds to $(Y\times Z)\cap \Delta$. Since $\Delta$ has dimension $n$, and since $r+s-n=(r+s)+n-2n$, we reduce to proving the result for the two varieties $Y\times Z$ and $\Delta$ in $\Bbb A^{2n}$. Now $\Delta$ is an intersection of exactly $n$ linear hypersurfaces, namely $x_i-y_i=0$ for $1\leq i \leq n$ where $x_1,\cdots,x_n,y_1,\cdots,y_n$ are the coordinates on $\Bbb A^{2n}$. Now applying the special case above $n$ times, we get the desired result.
Here are the relevant results referenced in this proof:
Krull's Principal ideal theorem: If $R$ is a noetherian ring and $I$ is a principal proper ideal, then all minimal primes over $I$ have height at most one.
Dimension theorem: Let $k$ be a field and let $B$ be an integral domain finitely generated as a $k$-algebra. Then for any prime ideal $\mathfrak{p}\subset B$, we have $\operatorname{height} \mathfrak{p} + \dim B/\mathfrak{p} = \dim B$.
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Is the proposition assuming that the intersection is non-empty? – Johnny T. Nov 03 '19 at 15:04
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1No. The intended meaning is that if the intersection is nonempty, then every irreducible component has the relevant dimension. If the intersection is empty, then as there aren't any irreducible components of the empty set, the statement is vacuously true. But this is no matter - the proposition exactly applies to your situation. – KReiser Nov 03 '19 at 17:34
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Sorry to bother after so long, but what about the example of 2 double cones in $\mathbb R^3$ intersecting only at the origin (the tip of both double cones)? Surely each double cone is 2 dimensional, and so this dimension formula would give the intersection to have dimension $\geq 1$. I'm very sorry if this is a stupid question; I'm very new to the subject. – D.R. Feb 08 '24 at 18:17
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1@D.R. the answer to this depends a little bit on what your foundations and assumptions are. If you're in the place where a variety is a subset of $\Bbb R^n$ cut out by polynomial equations, then this theorem doesn't apply (it assumes an algebraically closed base field). If you're using schemes, then the intersection is actually hiding in the complex world - $V(x^2+y^2-z^2,2x^2+2y^2-z^2)=V(x^2+y^2,z^2)$, for instance. – KReiser Feb 08 '24 at 19:24
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@KReiser so in the theorem you write about in your answer should $k$ be algebraically closed? – D.R. Feb 09 '24 at 00:16
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1@D.R. if you're not interpreting them as schemes, yes. I've made a small edit to try and make it a little clearer. – KReiser Feb 09 '24 at 01:17