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This question comes from Georges Elencwajg's wonderful answer to Why learning modern algebraic geometry is so complicated?

Let $ k $ be an algebraically closed field and consider the closed subset $ X $ of $ \mathbb{A}^4_{k} $ which is the union of the two planes $ x = y = 0 $ and $ z= w = 0 $. This is an algebraic set defined by the four polynomials $ xz, xw, yz, yw $. Why is it impossible for $ X $ to be the set-theoretic intersection of three hypersurfaces in $ \mathbb{A}^4_{k} $?

If one prefers schemes, then the question can be rephrased as: Let $ X \subset \mathbb{A}^{4}_{k} $ be the closed subscheme defined by the ideal $ I = (xz,xw,yz,yw) $ i.e., $ X = \operatorname{Spec} (k[x,y,z,w]/I) $. Is $ X $ the scheme-theoretic intersection of three hypersurfaces of $ \mathbb{A}^4_{k} $?

I hope I've got the question correct. I have no ideas on how to even start as the (very little) intersection theory I know is on projective spaces.

Cranium Clamp
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First, $X$ can be written set-theoretically as the intersection $$ xz = yw = xw - yz = 0 $$ of three hypersurfaces. This is easy to check.

But it is, indeed, cannot be written scheme-theoretically as such an intersection. The obstruction is that $$ I_X \otimes \mathcal{O}_o \cong \mathcal{O}_o^{\oplus 4}, $$ where $o \in \mathbb{A}^4$ is the origin. To deduce the above isomorphism one can note that $$ I_X \cong I_{X_1} \otimes I_{X_2}, $$ where $X_i$ are the planes (such that $X = X_1 \cup X_2$) and use the Koszul resolutions $$ 0 \to \mathcal{O} \stackrel{(y,-x)}\to \mathcal{O} \oplus \mathcal{O} \to I_{X_1} \to 0 $$ and $$ 0 \to \mathcal{O} \stackrel{(w,-z)}\to \mathcal{O} \oplus \mathcal{O} \to I_{X_2} \to 0 $$ to compute the tensor products.

Sasha
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  • Sorry, I don't quite understand the meaning of $ I _X \otimes O_o \cong O_o^{\oplus 4} $. Isn't $ I_X $ the ideal sheaf of $ X $ and $ O_o $ the local ring at the origin $ o $? – Cranium Clamp Feb 26 '20 at 20:27
  • No, this is not the local ring, this is the structure sheaf of the point (i.e., the quotient of the local ring by its maximal ideal). – Sasha Feb 27 '20 at 13:29
  • Oh, I now understand, thank you. I was just confused by the notation. Tensoring one of those exact sequences, say the first gives $ \mathcal{O}o \rightarrow \mathcal{O}_o \oplus \mathcal{O}_o \rightarrow I{X_1} \otimes \mathcal{O}o \rightarrow 0 $, where the first map $ (y, -x) $ now becomes the zero map. So we get $ I{X_1} \otimes \mathcal{O}_o \equiv \mathcal{O}_o \oplus \mathcal{O}_o $ and hence the obstruction... – Cranium Clamp Feb 28 '20 at 23:41