In Mumford's algebraic geometry I(page 37-38), he showed that if $S\subset \mathbb{C}^{m+1}$ is a subvariety and $S_{0}\subset S$ is Zariski open, then $p_{2}(S_{0})$ contains a Zariski-oepn subset of $\overline{p_{2}(S)}$. Here $p_{2}$ is the projection to the $\mathbb{C}^{m}$coordinate. He used the following argument during his proof:
Let $T=\overline{p_{2}(S)}$, then $T$ is a variety and the affine rings of $S$ and $T$ satisfy $R_{S}=\mathbb{C}[X_{1},...,X_{m+1}]/I(S)\leftarrow R_{T}=\mathbb{C}[X_{2},...,X_{m+1}]/I(T)$ Hence we have $R_{S}\cong R_{T}[X_{1}]/\mathscr{U}$ for some ideal $\mathscr{U}$. Mumford argued that assuming $\mathscr{U}\not=0$, this showed $\dim(S)=\dim(T)$ But why this is true? Checking definition give me $\dim(X)=\text{tr.d.}_{\mathbb{C}}\mathbb{C}(X)$, and it is not clear to me why the transcendental degree of two fields must be equal. This question is quite elementary so I venture to ask in here.
My other question is simpler:why it is suffice to prove the above statement to show the projection of constructable sets are constructable(in Zariski topology, of course)?