# Projective varieties are the common zeros of some homogeneous polynomials

definition: A projective variety over a field $k$ is a closed subscheme (over $k$) of $\mathbb P^n_k=\operatorname{Proj}(k[T_0,\ldots,T_n])$.

Now it can be proved (I have done it) that if $I=(f_1,\ldots,f_m)$ is an homogeneous ideal of $k[T_0,\ldots,T_n]$, then the projective algebraic set $Z_+(f_1,\ldots,f_m)\subseteq\mathbb P^n(k)$ is in bijection with the set of $k$-rational points of $\operatorname{Proj}\frac{k[T_0,\ldots,T_n]}{I}$. Pratically a scheme of the type $\operatorname{Proj}\frac{k[T_0,\ldots,T_n]}{I}$ corresponds to the naive intuition of an algebraic variety; now I can't find a proof of the following statement:

Every closed subscheme (over $k$) of $\mathbb P^n_k$ is isomorphic to a scheme $\operatorname{Proj}\frac{k[T_0,\ldots,T_n]}{I}$ for certain homogeneous ideal $I\subseteq k[T_0,\ldots,T_n]$.

So firstly I ask any hint, or complete proof or even a reference of the above proposition. Then I have the following question:

Is it true that $\operatorname{Proj}\frac{k[T_0,\ldots,T_n]}{I}$ is irreducible as scheme if and only if $f_1,\ldots,f_m$ are irreducible polynomials? (For affine schemes this is true)

Your last claim is not true "even" in the affine case. For example the affine plan set cut by the ideal $(x,(y-1)^2-x)$ is reducible yet the generators of the ideal are irreducible. You can construct a similar example in the projective plane by intersecting a line and a conic.