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Let $X,Y$ be two affine varieties and $m_x$ is the maximal ideal of the local ring $\mathcal{O}_{X,x}$; $m_x$ is the maximal ideal of the local ring $\mathcal{O}_{Y,y}$; $m_{x,y}$ is the maximal ideal of the local ring $\mathcal{O}_{X\times Y,(x,y)}$. Assume the dimension of $m_x/m_x^2$ is $N_x$, the dimension of $m_y/m_y^2$ is $N_y$, the dimension of $m_{x,y}/m_{x,y}^2$ is $N_{x,y}$. Then do we have $N_x+N_y=N_{x,y}$?

6666
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    $m_x/m_x^2$ is a quotient of an $\mathcal{O}{X,x}$-module $m_x$ by an $\mathcal{O}{X,x}$-submodule $m_x^2$ and so is an $\mathcal{O}_{X,x}$-module in its own right. – Angina Seng Sep 15 '19 at 19:15
  • I believe we have a short exact sequence (writing tildes for quotient) $0\to\tilde{m}x\to\tilde{m}{x,y}\to\tilde{m}y\to0$. The first map is inclusion, and the second kills all terms involving $X_1,\dots,X_n$. Assuming this, we can use the fact that the alternating sum of dimensions in a short exact sequence of vector spaces is $0$. In other words, $N_x-N{x,y}+N_y=0,$ which is what we needed. I am not an expert though so only posting as a comment. – Douglas Molin Sep 18 '19 at 08:34

1 Answers1

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Let $(A, m_A)$, and $(B, m_B)$ be the local rings corresponding to points $x$ and $y$ respectively. We will assume that $X, Y$ are defined over an algebraically closed field $k$. Let $C = A \otimes_k B$. Then we have the following equality

$$\Omega_{C/k} \cong \Omega_{A/k} \otimes_A C \oplus \Omega_{B/k} \otimes_B C.$$

The above equality follows from the universal property of differentials. Let $m_C = m_A \otimes_k B + A \otimes_k m_B$ be the maximal ideal in $C$ corresponding to the tuple $(x, y)$. Let $S = C \setminus m_C$ be the multiplicatively closed set. Since $\Omega_{S^{-1}C/k} \cong S^{-1}\Omega_{C/k}$ as $S^{-1}C$ modules, inverting $S$ in the above equation we get,

$$\Omega_{S^{-1}C/k} \cong \Omega_{A/k} \otimes_A S^{-1}C \oplus \Omega_{B/k} \otimes_B S^{-1}C.$$

Now, tensoring this above equation with $S^{-1}C/S^{-1}m_C \cong k$ in the above equality we have,

$$\Omega_{S^{-1}C/k} \otimes_{S^{-1}C} k\cong \Omega_{A/k} \otimes_A k \oplus \Omega_{B/k} \otimes_B k.$$

The above equality is true because of the following series of isomorphisms

$$(\Omega_{A/k} \otimes_A S^{-1}C) \otimes_{S^{-1}C} S^{-1}C/S^{-1}m_C \cong \Omega_{A/k} \otimes_A(S^{-1}C \otimes_{S^{-1}} S^{-1}C/S^{-1}m_C) \cong \Omega_{A/k} \otimes_A k.$$

and similarily for $B$. Now Proposition 8.7 from $\S$ 8 of Chapter 2 in Algebraic Geometry by R. Hartshorne, we have

$$\Omega_{A/k} \otimes_A k \cong m_A/m^2_A, \; \; \Omega_{B/k} \otimes_B k \cong m_B/m^2_B, \;\; \Omega_{S^{-1}C/k} \otimes_{S^{-1}C} S^{-1}C/m_C \cong m_C/m^2_{C}.$$

Thus we are done by comparing the dimensions in the second to last equation. Note that all we require is that the residue field of local rings are $k$.

Remark : The equality established above in particular implies that the product of smooth varieties is smooth.

random123
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  • Why is $S^{-1}C/S^{-1}m_C\cong k$ ? – 6666 Sep 19 '19 at 16:22
  • @6666 That is just a general statement. Let A and B be two $k$-algebras and $I, J$ be two ideals, then $A/I \otimes_k B/J$ is isomorphic to $(A \otimes_k B)/ (I \otimes_k B + A \otimes_k J)$. – random123 Sep 19 '19 at 16:48
  • Sorry how is that related to my question? why do you need two $k$-algebras for the $S^{-1}C/S^{-1}m_C$? – 6666 Sep 19 '19 at 18:54
  • @6666 $C$ is a tensor product of two algebras and $C/m_C$ is isomorphic to $k$. – random123 Sep 20 '19 at 02:06