Let $X\subset \mathbb{A}^n_K$ and $Y\subset \mathbb{A}^m_K$ be affine varieties. How can I prove that dimension of the product variety $X\times Y \subset \mathbb{A}^{m+n}_K$ is dim$X$+dim$Y$?
Here I am using the definition of dimension (say, of $X$) as the length of the maximal chain of prime ideals inside the ideal associated with $X$. However it seems that the formula dim$R$=trdeg$_K$ Quot$R$ is more of use here. Pursuing the approach of this formula with transcendental degree, I can show that: if $\beta$ and $\gamma$ are transcendental basis of $I(X)$ and $I(Y)$ respectively, then $I(X \times Y)$ is algebraic over $K(\beta, \gamma)$, but I cannot show that $\beta \cup \gamma$ is algebraic independent over $\mathbb{A}^{m+n}_K$. If I can show this then I'll be done.
Any help is appreciated!
Best Answer
Here is just an outline.
Consider the ring extension $k \to A(X)$. Use Noether normalization to see $A(X)$ is integral over a subring isomorphic to $k[x_1, \dots, x_s]$, where $s$ is the dimension of $X$ (using the fact that definition of dim $X$ in terms of chains of irreducible closed subsets is the same as the Krull dimension of the coordinate ring $A(X)$). Do the same for $k \to A(Y)$ to produce another subring $k[y_1, \dots, y_t]$, where $t=dim Y$.
Next show that coordinate ring of $X \times Y$ is integral over a subring isomorphic to $k[x_1, \dots, x_s, y_1, \dots, y_t]$. Finally use the lying over and going up theorems to conclude that $A(X \times Y)$ also has dimension $s+t$.