If you have two projective varieties $X$ and $Y$ such that $\dim(X) < \dim(Y)$, is it necessarily true that we can't have surjective morphism $f : X \to Y$? I am familiar with first chapter of Hartshorne.
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Yes, and the appropriate proof depends on your level of technology - for instance, one proof of the statement is this MSE question which is potentially inappropriate for certain levels or styles of learning algebraic geometry. Can you add more details about what you know and what technology you're familiar with, so answerers can give you an appropriate solution? – KReiser Apr 27 '22 at 22:47
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I am familiar with first chapter of Hartshorne. – Dhruv Apr 27 '22 at 22:55
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In the context of Hartshorne chapter I, the dimension of a variety $X$ over $k$ is the transcendence degree of its function field $k(X)$ over $k$. A dominant (in particular, a surjective) map of varieties $X\to Y$ induces a pullback map on function fields $k(Y)\to k(X)$. But given any sequence of maps of fields $k\to F\to E$, we have $\operatorname{trdeg} E/k = \operatorname{trdeg} E/F + \operatorname{trdeg} F/k$, and therefore $\operatorname{trdeg} E/k \geq \operatorname{trdeg} F/k$. So for any dominant morphism of varieties $X\to Y$ over $k$, we see that $\dim X\geq\dim Y$.
KReiser
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