Let $X$ be a scheme (for sake of simplicity say proper $k$-scheme) which is projective, therefore there exist a closed immersion $i_X: X \to \mathbb{P}^n_k$.
Let $f: N \to X$ be the normalization morphism of $X$.
Why and how to derive that $Y$ is then also projective?
We have to construct a closed immersion $i_N \to \mathbb{P}^m$?
One way might be a technical overkill: Thereis a criterion that a proper $k$-scheme $X$ is projective iff it containes an ample sheaf $L$. So since $f$ is finite we can pullback $L$ to $f^*L$ which stays ample, tada!
But I'm looking for a more intuitive /constructive argument in order to understand the geometry behind this fact. Is there a way to construct such a closed embedding $i_N \to \mathbb{P}^m$ explicitely using the local properties (so on ring level) of the normalization map. Are the numbers $n$ and $m$ related?