Vakil's Foundations of Algebraic Geometry, Exercise 5.5.E Exercise 5.5.E
Let $A = \sum_{n\in\mathbb{Z}} A_n$ be a graded commutative ring with a $\mathbb{Z}$ type grading.
Let $f \in A_d, d > 0$.
Suppose $f$ is invertible.
Let h-$Spec(A)$ be the set of homogeneous prime ideals of $A$.
Let $\psi\colon$ h-$Spec(A) \rightarrow Spec(A_0)$ be the map defined by $\psi(P) = P \cap A_0$.
Then $\psi$ is a bijection.
His hint is as follows.
Let $P_0 \in Spec(A_0)$.
Define $Q_n = \{x \in A_n| x^d/f^n \in P_0\}$.
Let $P = \sum_{n\in \mathbb{Z}} Q_n$.
Show that $x \in Q_n$ if and only if $x^2 \in Q_{2n}$.
Show that if $x, y \in Q_n$, then $x^2 + xy + y^2 \in Q_{2n}$ and hence $x + y \in Q_n$.
Then show that $P$ is a homogeneous ideal.
And then show that $P$ is prime.
I don't understand why $x^2 + xy + y^2 \in Q_{2n}$ if $x, y \in Q_n$ and hence $x + y \in Q_n$.
 A: You want to show that $(x+y)^2 = x^2 + 2xy + y^2$ is in $Q_{2n}$.  To do that, you must show that $(x+y)^{2d}/f^{2n} \in P_0$.  (You should note that $d = \deg(f)$.)  Expand $(x+y)^{2d}$ using the binomial theorem.  Every term will be of the form ${2d \choose i} x^i y^{2d-i}$.  In particular, either the exponent of $x$ or of $y$ will be at least $d$.  Use this plus the fact that $P_0$ is an ideal to conclude.
A: First, you missed a $2$ there, it should say 'then $x^2+2xy+y^2\in Q_{2n}$'. Now, suppose $x,y\in Q_n$. Then 
$$x^d/f^n,y^d/f^n\in P_0$$
Since $P_0$ is an ideal consisting of degree-0-elements and $f$ is supposed to be invertible, you can multiply with terms such as $x^jy^{d-j}/f^n$ and get new elements in $P_0$ of the form
$$x^ky^{2d-k}/f^{2n}$$
Then expand $(x^2+2xy+y^2)^d/f^{2n}$ with the multinomial theorem. You will see that each term is (up to scalar multiple) in the form above, therefore this is in $P_0$, too. We can conclude, that $x^2+2xy+y^2\in Q_{2n}$.
As you know,
$$x^2 + 2xy + y^2 = (x+y)^2$$
and since you have already proven $x^2\in Q_{2n}\Leftrightarrow x\in Q_n$, you get $(x+y)\in Q_n$.
