Radical of prime ideal in homogeneous localization is prime 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. 
 A: 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!
