I am working in the following problem from my algebraic geometry course:
Let $X$ be a projective set of $\mathbb{P}^n$. Prove that there exists a finite regular morphism $F:X\to\mathbb{P}^k$, where $k=\dim X$.
(In case of doubt finite means that $k[X]$ is integral over $F^*(k[\mathbb{P}^k])$ and $F(X)$ is dense in $\mathbb{P}^n$, which is equivalent with $F^*$ injective)
Any help or hint will be appreciated, thanks :)
Edit: Here is my attempt:
If $X=\mathbb{P}^n$ we are done, since we can take $F$ to be the identity. If $X\neq\mathbb{P}^n$ there exist $x\in\mathbb{P}^n\setminus X$, so we can consider the map $f_1$ as the maps that projects $X$ from $x$. If $f_1(X)=\mathbb{P}^{n-1}$ we can take $F=f_1$. If $f_1(X)\neq\mathbb{P}^{n-1}$ there exist $y\in\mathbb{P}^{n-1}\setminus f_1(X)$, so we can consider the map $f_2$ as the maps that projects $f_1(X)$ from $y$. If $f_2(f_1(X))=\mathbb{P}^{n-2}$ we can take $F=f_2\circ f_1$,....
Since this process will end when we arrive $\mathbb{P}^k$ we can say that $F=f_{n-k}\circ\dots\circ f_{1}$. Of course this $F$ is well defined and it must be the solution to the problem, but I don't know how to prove that it is finite. If I can show that $f_i$ (a projection from a point) is a finite regular morphism we will be done, but I don't have ideas for this.
Edit: I know @KReiser said it could be a duplicate, but there the answer just work for characteristic $0$ and it gives us a 'geometric solution' that i don't like so much. I would like to have a pure algebraic solution and, if possible, an explicit formula for $F$ or a constructive method.
Here is the more algebraic solution you're asking for. First, we should require that our base field is infinite - we could bypass this requirement if we were working in the affine/non-graded case (see for instance a very nice proof by Mel Hochster here which works over any field and in fact generalizes to arbitrary domains), but the structure of the argument we're using here does require this.
Let $k$ be an infinite field and let $R$ be a graded $k$-algebra which is finitely generated by $x_0,\cdots,x_n$ in degree one. Then there are $y_0,\cdots,y_m\in R$ so that each $y_i$ is a $k$-linear combination of the $x_j$, the $y_i$ are independent, and $R$ is finite over $k[y_0,\cdots,y_m]$. We prove this by induction on $n$: the base case where $R$ is generated by $x_0$ is clear, as either $R$ is a polynomial ring in $x_0$ or $R$ is finite over $k$.
For the inductive step, let $\varphi:k[x_0,\cdots,x_n]\to R$ be the obvious graded surjection. If $\ker\varphi=0$, we're done. Else, let $P\in \ker\varphi$ be a nonzero homogeneous element of degree $d$. I claim that up to a change of variables of the form $x_i=x_i+\lambda_ix_n$ and $x_n=\lambda_nx_n$, we may assume $P=x_n^d+\cdots$: writing $P$ as a polynomial in $x_0$ and expanding out after this change of variables, we have $$P=P(\lambda_0,\cdots,\lambda_n)x_n^d+\cdots$$ where the $\cdots$ represents terms of degree less than $d$ in $x_n$. As any nonzero polynomial over an infinite field has a nonzero value, we can find $\lambda_i$ so that $P(\lambda_0,\cdots,\lambda_n)\neq 0$ and therefore $k[x_0,\cdots,x_n]/(P)$ is finite over $k[x_0,\cdots,x_{n-1}]$ and surjects on to $R$. Now consider the image of $\varphi(k[x_0,\cdots,x_{n-1}])$ in $R$: by the previous sentence, $R$ is finite over this, and by induction, we can find an appropriate selection of $y_i$ so that $\varphi(k[x_0,\cdots,x_{n-1}])$ is module-finite over $k[y_0,\cdots,y_m]$. As the composition of finite extensions is finite, we're done.