Let $B$ be a graded ring, $B=\oplus_{d\ge 0} B_d$. If $f\in B$ is homogeneous, we let $B_{(f)}$ denote the subring of $B_f$ made up of elements of the form $af^{-N}$, $N>0$, where $a$ is a homogeneous element with $\deg a=N\deg f$. This is the homogeneous localization. Note that $B_f$ is a $B_{(f)}$-graded algebra.
While reading about the construction of projective schemes, I have encountered the claim that if $\mathfrak q$ is a prime ideal of $B_{(f)}$, then $\sqrt{\mathfrak q B_f}$ is a homogeneous prime ideal of $B_f$.
How does one prove this? It is supposedly easy once one makes the reduction to considering only homogeneous elements when testing the primality condition, but I do not see how to make that reduction, nor how to continue.
Very sketchy post since the argument is long and I've got to run.
The first thing is to prove that a homogeneous ideal $I$ is prime if and only if $xy \in I$ with $x, y$ homogeneous implies $x, y \in I$. The "only if" direction is the interesting one. I'll just give the idea of the argument: write out general elements as sums of their homogeneous components $x = x_d + x_{d - 1} + \cdots$ and $y = y_e + y_{e - 1} + \cdots$; if $xy \in I$ then $x_dy_e \in I$. Keep going.
It should be clear that $\sqrt{\mathfrak{q}B_f}$ is homogeneous: $\mathfrak{q}B_f$ is generated by degree $0$ elements and the radical of a homogeneous ideal is homogeneous. The proof of this last fact is similar to the proof above. It's a little annoying to check that it's prime but we can do it.
Just to simplify the notation: it's enough to check that if I have $x, y \in B_f$ homogeneous of degrees $d, e$ with $xy \in \mathfrak{q}B_f$ then one of $x, y$ is in $\sqrt{\mathfrak{q}B_f}$. Let $k$ be the degree of $f$. If $xy \in \mathfrak{q}B_f$ then $(xy)^n/f^{d + e} = (x^n/f^d)(y^n/f^e) \in \mathfrak{q}B_f \cap B_{(f)} = \mathfrak{q}$. Remember that $\mathfrak{q}$ is prime!