How can every affine scheme fit inside Z² = 0?

Throughout, an affine scheme means a finitely presented affine scheme over a field $k$. An affine-linear section is a scheme-theoretic intersection with a closed subscheme cut out by affine-linear equations. Write $M_N$ for the affine space of $N\times N$ matrices.

The construction will replace the equations of $X$ by affine-linear conditions on a matrix $Z$ satisfying $Z^2=0$. A fixed identity block forces maximal square-zero rank, placing the matrices in one smooth nilpotent orbit over a field, while the affine-linear section cuts out $X$ itself.

The algebraic construction works over an arbitrary commutative ring $R$; after constructing the section, we specialize to $R=k$ for the orbit and symplectic interpretation. For $a,b\in R$,

$$\begin{pmatrix}0&a\\ b&0\end{pmatrix}^{\!2} = \begin{pmatrix}ab&0\\0&ab\end{pmatrix}.$$

Affine-linear maps put $a,b$ into the off-diagonal entries and read either diagonal entry after squaring. Thus one $2\times2$ matrix square computes multiplication over every characteristic.

Size two is necessary for a characteristic-independent construction. In characteristic two, every scalar expression built from affine maps and unary squaring is a sum of Frobenius powers and has no mixed monomial $xy$.

Write

$$X=\operatorname{Spec}R[t_1,\ldots,t_n]/(f_1,\ldots,f_r)$$

and choose a straight-line circuit for the $f_i$, using constants, addition, subtraction and multiplication. A straight-line circuit introduces each new wire once and only uses earlier wires.

For example, the cusp $y^2=x^3$ may be written

$$u=x^2,\qquad v=ux,\qquad w=y^2,\qquad w-v=0.$$

Reserve one $1\times1$ diagonal block of a matrix $A$ for every scalar wire. For every multiplication gate $w_\ell=w_iw_j$, reserve a private block

$$\begin{pmatrix}0&w_i\\w_j&0\end{pmatrix}.$$

Set every unused entry of $A$ to zero and introduce a second matrix $B$. We will intersect the resulting affine-linear conditions with the graph $B=A^2$.

By the $2\times2$ identity, a diagonal entry of the private block of $B$ is $w_iw_j$; identify that entry affinely with $w_\ell$. Constants, copying, addition, subtraction and the final zero-output conditions are affine-linear. Together these conditions cut out an affine-linear closed subscheme $L_X\subset M_N\times M_N$.

Size. If the circuit has $s$ scalar wires and $m$ multiplication gates, this construction uses $N=s+2m$.

If

$$\Gamma_N=\{(A,B):B=A^2\},$$

then

$$X\cong L_X\cap\Gamma_N.$$

The isomorphism is scheme-theoretic. After eliminating $B$, the coordinate ring is the polynomial ring in the circuit wires modulo the gate and output equations. Every internal gate equation is monic in a fresh wire. Eliminating the wires in circuit order leaves exactly $R[t_1,\ldots,t_n]/(f_1,\ldots,f_r)$, including its nilpotents, embedded components and multiplicities.

Now send the graph of squaring into one square-zero equation by the affine-linear map

$$\iota_N(A,B)= \begin{pmatrix}A&I_N\\-B&-A\end{pmatrix}.$$

Put $E=A^2-B$. Direct block multiplication gives

$$\iota_N(A,B)^2 = \begin{pmatrix} E&0\\ AB-BA&E \end{pmatrix}, \qquad AB-BA=EA-AE.$$

The two diagonal blocks are $E$, while the lower-left block belongs to the ideal generated by the entries of $E$. Therefore the pullback of the ideal generated by the entries of $Z^2$ is exactly the graph ideal generated by the entries of $A^2-B$:

$$\iota_N^{-1}\!\left(\{Z\in M_{2N}:Z^2=0\}\right)=\Gamma_N$$

scheme-theoretically. Already this is an affine-linear section of the square-zero matrix scheme.

Let

$$\mathcal C_N= \left\{ \begin{pmatrix}A&I_N\\-B&-A\end{pmatrix}:A,B\in M_N \right\}.$$

The fixed top-right block forces rank at least $N$. A square-zero endomorphism of a $2N$-dimensional vector space satisfies $\operatorname{im}Z\subseteq\ker Z$, hence has rank at most $N$. Thus every field-valued point of $\mathcal C_N$ with $Z^2=0$ has maximal possible square-zero rank $N$.

Let

$$J_N=\begin{pmatrix}0&I_N\\0&0\end{pmatrix}, \qquad \mathcal O_N=\operatorname{GL}_{2N}\!\cdot J_N.$$

The orbit $\mathcal O_N$ has exactly the matrices with $Z^2=0$ and $\operatorname{rank}Z=N$. Indeed, for such a matrix,

$$\operatorname{im}Z=\ker Z=:K,$$

and the induced map $k^{2N}/K\to K$ is an isomorphism. Conversely, a subspace $K\in\operatorname{Gr}(N,2N)$ together with an isomorphism $k^{2N}/K\to K$ determines such a matrix. Hence $\mathcal O_N$ is the open isomorphism locus in the vector bundle

$$\operatorname{Hom}(k^{2N}/\mathcal S,\mathcal S) \longrightarrow \operatorname{Gr}(N,2N),$$

so it is smooth and irreducible, with

$$\dim \mathcal O_N=N^2+N^2=2N^2.$$

The cell needed below is the orbit of $J_N$ under a simple abelian unipotent subgroup. For $A\in M_N$, set

$$u(A)=\begin{pmatrix}I_N&0\\-A&I_N\end{pmatrix}.$$

Then, over every $k$-algebra,

$$u(A)\,J_N\,u(A)^{-1} = \begin{pmatrix}A&I_N\\-A^2&-A\end{pmatrix}.$$

As $A$ varies, $u(A)$ runs through an abelian unipotent subgroup isomorphic to $\mathbb G_a^{N^2}$. The conjugation identity identifies its orbit with

$$\mathcal U_N= \left\{ \begin{pmatrix}A&I_N\\-A^2&-A\end{pmatrix}:A\in M_N \right\} \cong\mathbb A_k^{N^2}.$$

It is closed in $\mathcal O_N$: inside $M_{2N}$ it is the graph cut out by

$$Z_{12}=I_N, \qquad Z_{22}=-Z_{11}, \qquad Z_{21}=-Z_{11}^2.$$

Combining the block-square and conjugation identities gives equalities of subschemes

$$\mathcal C_N\cap\{Z^2=0\} = \mathcal C_N\cap\mathcal O_N = \mathcal U_N.$$

These identities hold functorially on every $k$-algebra, hence as equalities of subschemes.

The same description holds for every $N$:

$$\ker Z(A)=\operatorname{im}Z(A)=\{(x,-Ax):x\in k^N\}=\operatorname{graph}(-A).$$

Thus $A\mapsto\ker Z(A)$ identifies $\mathcal U_N$ with the standard affine big cell of $\operatorname{Gr}(N,2N)$.

The identity block forces maximal square-zero rank, placing the cell in $\mathcal O_N$. As $A$ varies, $u(A)$ runs through an abelian unipotent subgroup whose orbit is $\mathcal U_N$. The trace pairing gives $\mathcal O_N$ its natural symplectic form; $\mathcal U_N$ is isotropic and half-dimensional, hence Lagrangian.