Dimension of the set of common zeros of $ (m+1) $ forms of equal degree Suppose we have a rational map $ \varphi: \mathbb{P}^{n}_{x_{0},\dots,x_{n}}  \dashrightarrow \mathbb{P}^{m}_{y_{0},\dots,y_{n}} $ such that $$ (x_{0},\dots,x_{n}) \longmapsto (F_{0}(x_{*}):\dots:F_{m}(x_{*})) \in \mathbb{P}^{m}_{y_{0},\dots,y_{m}},  $$
where the homogeneous polynomials $ F_{0}(x_{*}),\dots, F_{m}(x_{*}) $ are of equal degree $ d \geq 1, $ and have no common factors.
Then the set $$ X = \lbrace F_{0}(x_{*})=\dots=F_{m}(x_{*}) = 0 \rbrace \subset \mathbb{P}^{n}_{x_{0},\dots,x_{n}}, $$ is of dimension $ \leq n-2 $ where $ \varphi $ is not defined.
I don't see why this is true.
 A: This is because you're assuming the polynomials $F_i$ have no common factors.
This boils down to the following fact:

Every codimension 1 subvariety of $P=\mathbb{P}^n$ is of the form $V(F)$ for some homogeneous polynomial $F\in k[x_0,\dots,x_n]$.

Show the result assuming this fact and feel free to ask questions if you get stuck.
Let's show this fact. First we show the same result holds if we replace $P$ by $A=\mathbb{A}^n$. 
Proof of the affine case: Let $Z\subset A$ be an irreducible codimension 1 subvariety. In algebraic terms this translates to $I=I(Z)\subset k[X_1,\dots,X_n]$ being a height 1 prime ideal.
Note that $k[X_1,\dots,X_n]$ is an UFD. Hence $I$ is a principal ideal, so there exists some irreducible polynomial $f$ such that $I=(f)$ and $Z=V(f)$ and the claim follows for the affine case. This is the technical heart of the claim, see for example the Wikipedia article. 
Proof of projective case: We use the affine claim to show the projective one. Let $Z\subset P$ be a codimension 1 irreducible subvariety and let $\{U_i\}_{i=0,\dots,n}$ be the standard affine cover of $P$. There are two cases to distinguish:
1) $Z=V(x_i)$ is a "coordinate hyperplane", in which case we're done.
2) $Z$ is anything else. In particular, $Z\cap U_i\neq \emptyset$ for all $i=0,\dots, n$. Indeed, $Z\cap U_i=\emptyset$ implies $Z$ is contained in the $i$th coordinate hyperplane. Since both varieties have codimension 1 this would mean we're in case 1).
Then, for all $i$ the intersection $Z\cap U_i$ is a subvariety of dimension $n-1$ in the corresponding affine chart (since it's closed in there and it has the same dimension as in the ambient space). 
The affine result above implies that $Z\cap U_i=V(f_i)$ for some irreducible $f_i\in k[x_0/x_i,\dots,x_n/x_i]$. Call $F_i\in k[x_0,\dots, x_n]$ the homogeneization of $f_i$. 
To conclude we claim that $F_i=F_j$ for all $i\neq j$. Call this polynomial $F$. To see this you will need to strengthen our affine lemma above a little bit (hint: remember how two affine patches glue and use the fact that the localization of a UFD remains a UFD.)
This last step actually shows that $Z\cap U_i=V(F)\cap U_i$ for all $i$, so we're done.
