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I am not able to show explicitly if $X\rightarrow Y$ is a surjective morphism of affine algebraic varieties then the dimension of X is at least as large as the dimension of Y. Need geometric interpretation also. thank you

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up vote 1 down vote accepted

If we are talking maps between affine domains.

Let $A\to B$ be an inclusion of domains which are finitely generated $k$-algebras. Since dimension of affine domain is equal to its transcendent degree.

And $A\to B$ will give an injection $Q(A)\to Q(B)$, we win.

If just a map of finitely generated $k$-algebras, say $A\to B$ also.

Since the set map $\operatorname{Spec} B\to \operatorname{Spec} A$ is surjective, it follows that the kernel of $A\to B$ consists of nilpotents. So we may assume $A\to B$ is injective. Now the dimension of $A$ is equal to its transcendental degree, it gives that $\operatorname{dim}B\geq\operatorname{dim}A$.

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