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

Question

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

Answer 1

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.