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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.

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  • $\begingroup$ Isn't $\mathfrak{p} \oplus (1) \oplus (1) \oplus …$ prime for $\mathfrak{p} $ prime? $\endgroup$
    – user40276
    Aug 6, 2015 at 20:39
  • $\begingroup$ @user40276 That one is not even an ideal! $\endgroup$
    – user26857
    Aug 6, 2015 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, 2015 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, 2015 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, 2015 at 20:56

1 Answer 1

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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 $P_0=\mathfrak p_0$.

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