4
$\begingroup$

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 $N_x+N_y=N_{x,y}$?

$\endgroup$
  • 2
    $\begingroup$ $m_x/m_x^2$ 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. $\endgroup$ – Lord Shark the Unknown Sep 15 at 19:15
  • $\begingroup$ 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. $\endgroup$ – Douglas Molin Sep 18 at 8:34
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Why is $S^{-1}C/S^{-1}m_C\cong k$ ? $\endgroup$ – 6666 Sep 19 at 16:22
  • $\begingroup$ @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 $(A \otimes_k B)/ (I \otimes_k B + A \otimes_k J)$. $\endgroup$ – random123 Sep 19 at 16:48
  • $\begingroup$ Sorry how is that related to my question? why do you need two $k$-algebras for the $S^{-1}C/S^{-1}m_C$? $\endgroup$ – 6666 Sep 19 at 18:54
  • $\begingroup$ @6666 $C$ is a tensor product of two algebras and $C/m_C$ is isomorphic to $k$. $\endgroup$ – random123 Sep 20 at 2:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.