Dimension of $m/m^2$, where $m$ is the maximal ideal of $\mathcal{O}_{X\times Y,(x,y)}$

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

• $m_x/m_x^2$ 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. – Lord Shark the Unknown Sep 15 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 at 8:34

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.

• Why is $S^{-1}C/S^{-1}m_C\cong k$ ? – 6666 Sep 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 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 at 18:54
• @6666 $C$ is a tensor product of two algebras and $C/m_C$ is isomorphic to $k$. – random123 Sep 20 at 2:06