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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 $x + y \in Q_n$.

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2 Answers 2

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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.

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Thanks. I guess $x^2 + xy + y^2$(or $x^2 + 2xy + y^2$) is misleading. –  Makoto Kato Nov 15 '12 at 16:43

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 $(x+y)\in Q_n$.

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Your claim about which terms appear in the binomial theorem is not correct. You to use the fact that $\frac{x^i y^{d-i}}{f^n} \frac{x^d}{f^n} \in P_0$ and a similar fact with $y$ replacing $x$ in the second factor. –  Michael Joyce Nov 15 '12 at 16:34
    
Yup, thank you. I have fixed it. –  Gregor Bruns Nov 15 '12 at 16:45
    
A smaller nitpick is that I think your assertion is easier to see if you expand $(x+y)^{2d}$ with the simple binomial theorem than expanding $(x^2+2xy+y^2)^d$ using the multinomial coefficient version of the theorem (which is how I interpret your answer). –  Michael Joyce Nov 15 '12 at 16:47
    
It sure is. However, for the silly sake of tidiness, my solution follows the path the hint suggests (without shortcut). :) –  Gregor Bruns Nov 15 '12 at 16:53

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