I've got this exercise asking me to prove first that the product of quasi-projective varieties $X$ and $Y$ (henceforth just "varieties") is irreducible iff both $X$ and $Y$ are. I managed to solve this part. Secondly, I'm asked to prove that the dimension of the product is $n+m$ (where $n=dimX, m=dimY$). I guess I'll need to use the Segre embedding properly, or more precisely, it's image in $\mathbb{P}^{(n+1)(m+1)-1}$. Here's my problem: I really don't know how to compute this dimension, not even for the simplest case, the product variety $\mathbb{P}^{n} \times \mathbb{P}^{n} $
The definition of dimension we were given in this course is that of the transcendence degree of the field of rational functions of the variety, but I've only seen examples of computations in very simple cases, like that of an affine curve, where you can conclude dim is less than 2 because $x$ and $y$ are not algebraically independent sice there is a polynomial that identically vanishes on the curve, hence in its field of rational functions.
Any advice?? Thanks in advance.
Edit: I managed to prove that $dim(\mathbb{P}^n \times \mathbb{P}^m) \geq n+m$, using the chain of proper irreducible closed sets generated by the ideals $(x_1),(x_1, x_2),...,(x_1,...,x_n),(x_1,...,x_n,y_1),....,(x_1,...,x_n,y_1,...,y_m)$, (there is some proposition telling me that closed sets in the product are exactly the zero loci of bihomogeneous polynomials, i.e. homogeneous with respect to the $x_i$'s and to the $y_j$'s, separately. These are all bihomogeneous of bidegree (1,0) and (0,1)). A proposition then tells me that dimensions must be strictly increasing in this case (strictly ascending irreducible). But I really don't know how to bound it from above, nor how to prove it for more general varieties. I thought about using a theorem which says that any irreducible variety is birational to a hypersurface, but I don't even know if birationality is preserved by the product, nor how to proceed with this idea even if birationality were preserved.