How to show $\coprod^n_{i=1}X_i\to S$ induced by $f_i$ is projective? A morphism of schemes $f: X\to Y$ is projective if  $f$ can be factorized as $X \to \mathrm{P}_Y^n\to Y$ for some closed immersion $X\to \mathrm{P}_Y^n$.
Suppose $f_i: X_i\to S$ is projective for $i=1,2,\cdots,k$.
How to show $\coprod^k_{i=1}X_i\to S$ induced by $f_i$ is projective?
 A: We are looking for maps $X_i\to\mathbf{P}^{n_i}\to\mathbf{P}^m$ such that each the image of each $X_i$ are disjoint to each other. We might as well require that the image of each $\mathbf{P}^{n_i}$ are disjoint. There are many way to embed $n$ projective spaces into a bigger one such that they don't intersect each other. Over an infinite field, it is geometrically clear that you can do this if $m=2\operatorname{max}(x_i)+1$. In any case, you can take $m=\sum(n_i+1)-1$ with the map
$$\mathbf{P}^{n_i}\to\mathbf{P}^m, [x_0:...:x_{n_i}]\mapsto  [0:...:0:x_0:...x_{n_i}:0:...:0]$$
where the number of zeros before $x_0$ is $\sum_{j<i}(n_j+1)$.
A: Each $f_i: X_i\to S$ can be factorized as $X_i\to \mathrm P_S^{n_i}\to S$.
Let $m=\max\{n_i\}$, we have the canonical morphisms $\mathrm P_S^{n_i}\to \mathrm P_S^{m}$ which are closed immersions, then we have the composite morphisms of closed immersions $X_i\to \mathrm P_S^{n_i}\to\mathrm P_S^{m} $. 
Finally, we get the desired morphism $\coprod_{i=1}^k X_i\to \mathrm P_S^{m}\to S$.
Is it correct?
A: Each $f_i: X_i\to S$ can be factorized as $X_i\to \mathrm P_S^{n_i}\to S$.
Let $m=\Sigma_{i=1}^k\{n_i\}$, we have the canonical morphisms $\mathrm P_S^{n_i}\to \mathrm P_S^{m}$ which are closed immersions, then we have the composite morphisms of closed immersions $X_i\to \mathrm P_S^{n_i}\to\mathrm P_S^{m} $. 
Finally, we get the desired morphism $\coprod_{i=1}^k X_i\to \mathrm P_S^{m}\to S$.
Is it correct?
