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Hi! I am following the hint given in Exercise 4.5.E in Vakil's Foundations of Algebraic Geometry, but I am stuck trying to prove that if $a_1,a_2 \in Q_i$, then $a_1^2 + 2a_1 a_2 + a_2^2 \in Q_{2i}$.

What I have tried so far: We have $$a_1^2 + 2a_1 a_2 + a_2^2 \in Q_{2i} \text{ iff } (a_1^2 + 2a_1 a_2 + a_2^2)^{\deg f}/f^{2i} \in P_0.$$ I have tried expanding this using the binomial formula, but this doesn't get me anywhere (at least not at the moment) as I would like each term to be in $P_0$, however I can't see why this should be the case for any term except the first and last (i.e. the completely obvious ones). I am hoping for a very generous hint or even just a complete solution.

By the way this is not homework, or an assignment or anything like that, this is just me trying to get a better understanding of Algebraic Geometry by working through the exercises in Vakil's notes, as I find this very enlightening (both when it comes to rigour and intuition). Thanks in advance.

Added for the sake of self-containment: In this exercise we are going to show that if $S$ is a graded ring, then there is a bijection between the prime ideals of $(S_f)_{0}$ (the degree zero elements in $S$ localized at the element $f \in S$) and the homogeneous prime ideals in $S$ not containing $f$. We will avoid notation by proving the slightly stronger statement:

If $A$ is a graded ring with an invertible element $f$ of positive degree, then there is a bijection between the prime ideals of $A_0$ and the homogeneous prime ideals of $A$.

One direction is obvious (just use the inclusion $A_0 \to A$), for the other given a prime ideal $P_0$ in $A_0$, we define $P = \oplus Q_i$, where $Q_i \subset A_i$, and an element $a \in A_i$ is contained in $Q_i$ iff $a^{\deg(f)}/f^i \in P_0$. We are going to show that this is a prime ideal, and following the hints given, one of the steps in the proof of this claim is showing the above (i.e. $a_1^2 + 2a_1a_2+a_2^2\in Q_{2i}$). The entire exercise can be read on page 146 in the notes, here.

  • I don't see how $A = k[x,x^{-1}]$ should cause problems as we want a bijection between primes of $A_0$, which in this case is just $(0)$ and the homogeneous prime ideals of $A[x,x^{-1}]$. Now we know that giving a prime ideal in $A[x,x^{-1}]$, is the same as giving one in $A[x]$ not containing $x$. As $A[x]$ is a PID, the prime ideals are just the ones generated by irreducible polynomials i.e $P = (f)$, however $P$ must not contain $x$ and thus either $P = (0)$ or $P$ is not homogeneous as it will be some polynomial with a constant term + terms of higher degree. – Oliver E. Anderson Feb 28 '14 at 10:39
  • Closely related to http://math.stackexchange.com/questions/235253/hartshornes-proof-of-proposition-2-5-chapter-ii-of-his-book-algebraic-geometry and http://math.stackexchange.com/questions/54456/the-bijection-between-homogeneous-prime-ideals-of-s-f-and-prime-ideals-of-s – user26857 Mar 03 '14 at 13:38

1 Answers1

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If $A$ is a graded ring with an invertible element $f$ of positive degree, then there is a bijection between the prime ideals of $A_0$ and the homogeneous prime ideals of $A$.

Hint. The correspondence is given by $\mathfrak p_0\mapsto\sqrt{\mathfrak p_0A}$.

Edit. Although I consider the way Vakil proves the above claim as far from best, let's show that if $a_1,a_2\in Q_i$ then $(a_1+a_2)^2\in Q_{2i}$: set $r=\deg f$ and then we have $a_1^r/f^i\in P_0$, $a_2^r/f^i\in P_0$, and want to show that $(a_1+a_2)^{2r}/f^{2i}\in P_0$. This is obvious since for $k+l=2r$ and $k\ge r$ (if $k<r$ then $l>r$ and use a similar argument) we can write $a_1^ka_2^l/f^{2i}=(a_1^r/f^i)(a_1^{k-r}a_2^l/f^i)$ and note that $\deg(a_1^{k-r}a_2^l/f^i)=i(k-r)+il-ir=0$.

user26857
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    Thanks a lot :) This was obvious as you said, I just failed to think that $(a_1^2 + 2a_1a_2 + a_2^2)^{deg f} = ((a_1 + a_2))^{2deg f}$. – Oliver E. Anderson Mar 03 '14 at 18:32
  • How does this show $(a_1+a_2)^{2r}/f^{2i}\in P_0$? – user5826 Jan 23 '21 at 02:11
  • @user5826 $(a_1 + a_2)^{2r}/f^{2i}$ can be expanded by the binomial theorem, and the above post shows that each individual term is in $P_0$ so the original expression must also be in $P_0$ because $P_0$ is a (prime) ideal. – Kyaw Shin Thant Mar 03 '24 at 12:02