Dimension of an irreducible affine variety by transcendence degree
= Dimension of an irreducible affine variety by transcendence degree
For an irreducible affine variety $X$ over $k$, its dimension is
$$
\dim X=\operatorname{trdeg}_k k(X),
$$
where $k(X)=\operatorname{Frac}(k[X])$ is its <function field of an algebraic variety>. An inclusion $k(Y)\hookrightarrow k(X)$ therefore implies $\dim Y\leq\dim X$.