The homogeneous coordinate ring satisfies $\dim S(Y) = \dim Y + 1$

Cover \(Y\) by the affine charts \(Y_i\). Then \(\dim Y = \sup_i \dim Y_i\), and since there are only finitely many charts the supremum is attained; after relabelling, assume it is attained at \(Y_0\).

The chart ring is the degree-zero part of a localization. Define maps between \(A(Y_0)\) and the degree-zero piece \(\qty{S(Y)_{x_0}}_0\) by \[ \alpha(f) \da f\qty{1, \frac{x_1}{x_0}, \ldots, \frac{x_n}{x_0}}, \qquad \beta(g) \da x_0^{\deg g} \cdot g(x_0, x_1, \ldots, x_n) . \] These are mutually inverse ring maps: on the one hand \[ \alpha(\beta(g)) = x_0^{\deg g} \, g\qty{1, \frac{x_1}{x_0}, \ldots, \frac{x_n}{x_0}} = g(x_0, x_1, \ldots, x_n), \] and on the other hand \(\beta(\alpha(f)) = f\) by the same computation read backwards. Hence \(A(Y_0) \cong \qty{S(Y)_{x_0}}_0\).

Identifying the localization. We have \(S(Y) \cong A(Y_0)[x_0]\) and \(S(Y)_{x_0} \cong S(Y)[\inverseof{x_0}]\), so \[ S(Y)_{x_0} \cong A(Y_0)[x_0, \inverseof{x_0}] . \]

Counting transcendence degree. Adjoining \(\inverseof{x_0}\) adds nothing transcendental, since \(x_0\) and \(\inverseof{x_0}\) satisfy the relation \(uv - 1 = 0\). So the transcendence degree of \(A(Y_0)[x_0, \inverseof{x_0}]\) over \(A(Y_0)\) equals that of \(A(Y_0)[x_0]\) over \(A(Y_0)\), namely \(1\), and therefore \[ \trdeg_k A(Y_0)[x_0, \inverseof{x_0}] = \trdeg_k A(Y_0) + 1 . \] Using the isomorphism above, \(\trdeg_k S(Y)_{x_0} = \trdeg_k A(Y_0) + 1\).

Conclusion. Since \(\dim Y_0 = \trdeg_k A(Y_0)\) and \(\dim S(Y)_{x_0} = \trdeg_k S(Y)_{x_0}\), this reads \[ \dim S(Y)_{x_0} = \dim Y_0 + 1 = \dim Y + 1 . \] It remains to see \(\dim S(Y)_{x_0} = \dim S(Y)\), and this holds because \(x_0 \in S(Y)\) already, so that \(S(Y)_{x_0} \cong S(Y)[\inverseof{x_0}]\) adds only the element \(\inverseof{x_0}\), which is algebraically dependent on \(x_0\) via \(uv - 1\).