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    Products of affine varieties and the tensor product of coordinate rings

    P-AGH315AFFPRODUCT

    Area
    Algebraic Geometry
    Topics
    Coordinate Rings, Dimension, Products
    Status
    draft

    Let \(X \subseteq \AA^n\) and \(Y \subseteq \AA^m\) be affine varieties.

    1. Show that \(X \times Y \subseteq \AA^{n+m}\) with its induced topology is irreducible. The affine variety \(X \times Y\) is called the product of \(X\) and \(Y\). Note that its topology is in general not the product topology.

    2. Show that \(A(X \times Y) \cong A(X) \tensor_k A(Y)\).

    3. Show that \(X \times Y\) is a product in the category of varieties, that is:

    • the projections \(X \times Y \to X\) and \(X \times Y \to Y\) are morphisms, and

    • given a variety \(Z\) and morphisms \(Z \to X\) and \(Z \to Y\), there is a unique morphism \(Z \to X \times Y\) commuting with the projections.

    1. Show that \(\dim (X \times Y) = \dim X + \dim Y\).
    Hint

    For (a), suppose \(X \times Y = Z_1 \union Z_2\) with each \(Z_i\) closed. Let \(X_i = \ts{ x \in X \st \ts{x} \times Y \subseteq Z_i }\). Show that \(X = X_1 \union X_2\) with \(X_1, X_2\) closed, so \(X = X_1\) or \(X = X_2\), and hence \(X \times Y = Z_1\) or \(Z_2\).

    Solution

    Write affine coordinates on \(\AA^{n+m}\) as \((x_1,\ldots,x_n,y_1,\ldots,y_m)\).

    1. If \(Z\subseteq X\times Y\) is closed, then\[ X_Z=\{x\in X:\{x\}\times Y\subseteq Z\} \]is closed in \(X\).

    Choose finitely many polynomials \(F_1,\ldots,F_r\in k[x_1,\ldots,x_n,y_1,\ldots,y_m]\) whose common zero set on \(X\times Y\) is \(Z\). Fix one \(F=F_\ell\). Its image in \[ k[x_1,\ldots,x_n]\tensor_k A(Y) \] is a finite sum \[ \sum_{j=1}^N a_j(x)e_j, \] where the \(e_j\in A(Y)\) may be chosen \(k\)-linearly independent. For a point \(x\in X\), the polynomial \(F(x,-)\) vanishes at every point of \(Y\) exactly when its class in \(A(Y)\) is zero, which by linear independence is equivalent to \[ a_1(x)=\cdots=a_N(x)=0. \] Thus the condition that \(F\) vanish on the whole fiber \(\{x\}\times Y\) cuts out a closed subset of \(X\). Intersecting these closed subsets for \(F_1,\ldots,F_r\) gives \(X_Z\).

    2. The subset \(X\times Y\subseteq\AA^{n+m}\) is irreducible, proving (a).

    1.For each fixed \(x\in X\), the fiber \(\{x\}\times Y\) is isomorphic to the irreducible variety \(Y\).It is the union of the two closed subsets\[ Z_1\cap(\{x\}\times Y), \qquad Z_2\cap(\{x\}\times Y), \]so it is contained in one of the \(Z_i\).Hence \(X=X_1\cup X_2\).

    Step 1 makes \(X_1\) and \(X_2\) closed in the irreducible variety \(X\). Thus \(X=X_1\) or \(X=X_2\). If, say, \(X=X_1\), every fiber lies in \(Z_1\), so \(X\times Y=Z_1\). This is exactly irreducibility.

    3. The natural homomorphism\[ A(X)\tensor_k A(Y)\longrightarrow A(X\times Y) \]is an isomorphism, proving (b).

    Let \[ S=k[x_1,\ldots,x_n,y_1,\ldots,y_m]. \] The quotient map \[ S\longrightarrow A(X)\tensor_k A(Y) \] has kernel \[ I(X)S+I(Y)S. \] It therefore suffices to show that this kernel is exactly \(I(X\times Y)\).

    Every element of \(I(X)S+I(Y)S\) vanishes on \(X\times Y\). Conversely, let \(F\in I(X\times Y)\) and write its image in the tensor product as \[ \overline F=\sum_{j=1}^N a_j\tensor b_j \] with \(b_1,\ldots,b_N\in A(Y)\) \(k\)-linearly independent. For each \(x\in X\), vanishing of \(F\) on \(\{x\}\times Y\) says \[ \sum_j a_j(x)b_j=0 \] in \(A(Y)\). Linear independence gives \(a_j(x)=0\) for every \(j\) and every \(x\in X\). Each \(a_j\) is therefore the zero element of \(A(X)\), so \(\overline F=0\). Thus \[ I(X\times Y)=I(X)S+I(Y)S, \] and taking quotients gives the asserted isomorphism.

    4. The coordinate projections are morphisms, and the usual pairing of two morphisms gives the required universal morphism.

    The first projection \[ \pi_X:X\times Y\to X \] is given by the coordinate functions \(x_1,\ldots,x_n\), and the second by \(y_1,\ldots,y_m\); hence both are morphisms.

    Given morphisms \(f:Z\to X\) and \(g:Z\to Y\), define \[ h:Z\longrightarrow X\times Y, \qquad h(z)=(f(z),g(z)). \] Locally on \(Z\), the coordinate functions of \(f\) and \(g\) are regular, so all \(n+m\) coordinate functions of \(h\) are regular. Thus \(h\) is a morphism, and by construction \[ \pi_X\circ h=f, \qquad \pi_Y\circ h=g. \] Any map satisfying these two equations must send \(z\) to \((f(z),g(z))\), so \(h\) is unique. This proves (c).

    5. The dimension satisfies\[ \dim(X\times Y)=\dim X+\dim Y. \]

    Put \(A=A(X)\) and \(B=A(Y)\). By step 3, \[ A(X\times Y)=A\tensor_k B, \] which is a domain by step 2.

    Choose Noether normalizations \[ k[u_1,\ldots,u_r]\subseteq A, \qquad k[v_1,\ldots,v_s]\subseteq B, \] with \(A\) and \(B\) integral over the indicated polynomial rings. Here \[ r=\dim X, \qquad s=\dim Y. \] Because tensoring injections of \(k\)-vector spaces preserves injectivity, the tensor product of the two polynomial subrings embeds as \[ k[u_1,\ldots,u_r,v_1,\ldots,v_s] \subseteq A\tensor_k B. \] The latter ring is integral over this polynomial subring: generators coming from \(A\) and from \(B\) satisfy the same monic equations after tensoring. Integral extensions preserve Krull dimension [@AM18, Chapter 5], so \[ \dim(A\tensor_k B)=r+s=\dim X+\dim Y. \] This proves (d).

    6. Q.E.D.

    Steps <1>2–<1>5 prove parts (a)–(d), respectively.

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