Why $\dim(S)=\dim(T)$? 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)?
 A: Well, let's first figure out where the isomorphism $R_T[X_1]/U\to R_S$ comes from. The map $R_T\to R_S$ you are mentioning is actually a monomorphism (because $p_2(S)$ is dense in $T$). The canonical extension $R_T[X_1]\to R_S$ need not be too, but it is well defined and surjective (I assume you can figure this out by yourself). Then $U$ being the kernel of this morphism, we get an isomorphism $R_T[X_1]/U\to R_S$ as desired.
Therefore we have $\dim(T) \leq \dim(S) \leq dim(T) + 1$ and the latter inequality is strict (that's the claim) if and only if $X_1\in\mathbb{C}(S)$ is algebraic over $\mathbb{C}(T)$; but this information is exactly decoded by $U$: The composition $R_T\hookrightarrow R_T[X_1]\to R_T[X_1]/U \xrightarrow{\cong} R_S$ coincides with the morphism we started with. In particular it's injective, thus $U$ is either trivial or has an element of the form $ a_n X_1^n + a_{n-1}X_1^{n-1} + \dots + a_0 $ for some $a_i\in R_T$, $a_n\not=0$, but after passing to the fraction fields this is nothing but the algebraicity of $X_1\in Frac(R_T[X_1]/U)$ over $\mathbb{C}(T)$. Hence if $U$ is non-trivial, $\dim(S)=\dim(T)$.
