5
$\begingroup$

Let $B=\oplus_{n\in\mathbb Z} B_n$ be a graded ring (commutative with 1). We know that $B_0$ is a subring of $B$, so we have the inclusion $B_0\hookrightarrow B$.

My question is:

Is every prime ideal of $B_0$ the inverse image of some homogeneous prime ideal in $B$?

If $B_0$ is a Principal Ideal Domain, then it's true, so I was trying to find a counterexample in $\mathbb Z[t][x,y]$, but it's not so easy.

$\endgroup$
  • $\begingroup$ Isn't $\mathfrak{p} \oplus (1) \oplus (1) \oplus …$ prime for $\mathfrak{p} $ prime? $\endgroup$ – user40276 Aug 6 '15 at 20:39
  • $\begingroup$ @user40276 That one is not even an ideal! $\endgroup$ – user26857 Aug 6 '15 at 20:46
  • $\begingroup$ @user26857 Why? It's the inverse image of $\mathfrak{p}$ in the projection $\prod_i A_i \rightarrow A_0$. Oh! I just realized that the question is about the direct sum. $\endgroup$ – user40276 Aug 6 '15 at 20:49
  • $\begingroup$ But you can still pick the inverse image of the projection $\oplus_i B_i \rightarrow B_0$. $\endgroup$ – user40276 Aug 6 '15 at 20:55
  • $\begingroup$ $B_n$ aren't rings! there may exist $x\in B_1$ and $x^{-1}\in B_{-1}$, so $P\oplus(\oplus_{n\ne 0} B_n)$ isn't an ideal $\endgroup$ – Exodd Aug 6 '15 at 20:56
2
+50
$\begingroup$

Let $A$ be a $\mathbb Z$-graded commutative ring, and $\mathfrak p_0$ be a prime ideal of $A_0$. Then there is a graded prime ideal $P$ of $A$ such that $P_0=\mathfrak p_0$. (Here $P_0$ denotes degree zero component of $P$, that is, $P\cap A_0$.)

For each $n\in\mathbb Z$ set $P_n=\{a\in A_n:aA\cap A_0\subseteq\mathfrak p_0\}$. Now set $P=\bigoplus_{n\in\mathbb Z}P_n$. Then $P$ is a prime ideal of $A$ and $P_0=\mathfrak p_0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.