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Let $A=\oplus A_i$ be a graded ring. Let $\mathfrak p$ be a minimal prime in $A$. Is $\mathfrak p$ a graded ideal?

Intuitively, this means the irreducible components of a projective variety are also projective varieties. When $A$ is Noetherian, I can give a proof, as follows.

There is some filtration of $A$, as an $A$ module, $$0=M_0\subset M_1\subset\cdots\subset M_n=A$$

such that $M_i/M_{i-1}\cong A/\mathfrak p_i$, for some graded prime ideal $\mathfrak p_i$.

Then I claim that the nilradical is $\cap\mathfrak p_i$. This is because

$$x^n=0 \Rightarrow x^nA=0 \Rightarrow x^nM_i\subset M_{i-1}\forall i \Rightarrow x^n\in \cap \mathfrak p_i \Leftrightarrow x\in \cap \mathfrak p_i $$ and

$$ x\in \cap \mathfrak p_i \Rightarrow xM_i\subset M_{i+1},\forall i \Rightarrow x^nA=0 \Leftrightarrow x=0.$$ Hence the mininal primdes are just the minimal elements in $\{\mathfrak p_i\}$.

I would like to know if this assertion is still true if we drop the Noetherian condition, or if anyone has some more direct proofs.

Thanks!

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1 Answer 1

up vote 3 down vote accepted

Yes, the minimal primes of a graded ring are graded. If $\mathfrak{p}$ is any prime, then the ideal $\mathfrak{p}^h$ generated by the homogeneous elements of $\mathfrak{p}$ is also prime, and certainly graded. So if $\mathfrak{p}$ is minimal, $\mathfrak{p}=\mathfrak{p}^h$ is graded.

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Are sure! It's so succinct! Thanks! –  Andrew Jul 29 '12 at 0:40
1  
In fact all associated prime ideals are graded. –  user26857 Jul 30 '12 at 19:58

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