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I met the following problem when I studied graded ring theory. I have no idea to solve it. Please help me. Thank you very much !

Let $R$ be a commutative graded ring, $M$ is a graded R-module, $N$ is a submodule of $M$. Denote by $N^*$ for the submodule of $M$ generated by all the homogeneous elements contained in $N$. Prove that: rad$(Ann_{R}M/N^{*})$=(rad $Ann_{R}M/N)^{*}$

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$\DeclareMathOperator\rad{rad}\newcommand\qed{\square}\DeclareMathOperator\Ann{Ann}$I'm a little surprised this didn't get closed immediately as being non-research or even possibly homework, but since it is open, I want to present a more mechanized version of Andrew Davies' answer. It's literally the same proof, but highlights some lemmas that are, ultimately, a lot more useful than just the fact you are asking about.

Lemma 1: Let $R$ be a ring, $V_1$ and $V_2$ two $R$-modules, and let $V_1 \to V_2$ be a surjection. Then $\Ann{V_1} \subset \Ann{V_2}$. $\qed$

Lemma 2: Let $R$ be a ring, $I_1$ and $I_2$ two ideals, and let $I_1 \subset I_2$. Then $\rad{I_1} \subset \rad{I_2}$. $\qed$

Lemma 3: Let $R$ be a graded ring, $V_1$ and $V_2$ two graded $R$-modules, and suppose $V_1 \subset V_2$. Then $V_2 / V_1$ is graded. $\qed$

Lemma 4: Let $R$ be a graded ring, $V$ a graded $R$-module. Then $\Ann{V}$ is a graded (homogeneous) ideal.

Proof: Let $v_d \in V$ be a homogeneous element of degree $d$, $x = \sum x_i$ a sum of homogeneous elements of degree $i$ in $R$. Since $xv_d = 0$ and the $x_i v_d$ have distinct degrees $d + i$, they must all be zero. Since this is true for any $v_d$, we conclude that all the $x_i$ kill the homogeneous elements of $V$, and thus since it is graded they annihilate $V$, so $\Ann{V}$ is the sum of its homogeneous subideals. $\qed$

Lemma 5: Let $R$ be a graded ring, $I$ a graded ideal. Then $\rad{I}$ is a graded (homogeneous) ideal.

Proof: By considering $R/I$, we may suppose that $I = 0$. (By the way, this argument is the key fact that connects the "$x^n \in I$" and the "$\bigcap P$" definitions of the radical.) If $x^n = 0$, write $x = \sum x_i$ as a sum of homogeneous elements of degree $i$. We show by induction on the number $s$ of summands that all the $x_i$ are nilpotent. When $s = 0$, the claim is tautological. Otherwise, let $x_m$ be the highest-degree term, so $x_m^n$ is the unique highest-degree term of $x^n$, and is therefore zero. Since $x_m$ is nilpotent, so is $x - x_m$, which has $s - 1$ summands, all of which are nilpotent by induction. Therefore the nilradical is the sum of its homogeneous subideals. $\qed$

Lemma 6: Let $R$ be a graded ring, $V_1$ and $V_2$ $R$-modules with $V_2$ graded and $V_1 \subset V_2$, and define $V_1^*$ to be the submodule of $V_1$ spanned by its elements homogeneous in $V_2$. Then $V_1^*$ is the unique maximal homogeneous submodule of $V_1$. $\qed$

Now the solution to your problem is as follows. Since $N^* \subset N$ we have a surjection $M/N^* \to M/N$, so $\Ann(M/N^*) \subset \Ann(M/N)$ by Lemma 1 and then $\rad \Ann (M/N^*) \subset \rad \Ann(M/N^*)$ by Lemma 2. Since $N^*$ is a graded submodule, by Lemma 3 the quotient $M/N^*$ is graded, so by Lemma 4 so is $\Ann(M/N^*)$, so by Lemma 5 so is $\rad \Ann(M/N^*)$. By Lemma 6 it is contained in $(\rad\Ann(M/N))^*$. Conversely, if $x \in (\rad\Ann(M/N))^*$ is homogeneous, then $x^n M \subset N$. But $x^n M$ is a homogeneous submodule, so again by Lemma 6, we have $x^n M \subset N^*$, and thus $x \in \rad\Ann(M/N^*)$, as desired.

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I think I have a proof.

To show the right side is contained in the left, let $x \in R$ be a homogeneous element for which there exists $n \in \mathbb{Z}$ such that $x^nM \subset N$. Given any $m \in M$ break it up into homogeneous pieces, $m=m_1+\ldots+m_s$ say. Each $x^nm_i$ lies in a different homogeneous piece of $M$ by the graded structure and since $x$ is homogeneous. But the annihilator condition implies it also lies in $N$ and $N^{\ast}$ by definition - it is the submodule generated by all homogeneous elements. Thus $x$ belongs to the left hand side.

For the opposite inclusion, suppose that for some $x \in R$ there exists $n \in \mathbb{Z}$ such that $x^nM \subset N^{\ast}$. Since $N \supset N^{\ast}$ it is sufficient to show that each homogeneous piece of $x$ lies in the right hand side. I think induction on the number of homogeneous pieces is a good way to go. The base case is clear.

Let $x=x_1+x_2\ldots+x_s$ be a homogeneous decomposition where $x_1$ is the highest degree piece, with $(x_2 + \ldots +x_s)$ belonging to the right hand side (that is, each $x_i$ with $i \geq 2$ lies in the right hand side). Let $m \in M$ be a homogeneous element. Then $x^nm=x_1^nm + l.o.t$ where the other terms have smaller degree. Since $x^nm \in N^{\ast}$ each homogeneous piece must be too. $x_1^nm$ lies in a higher degree piece than everything else and so this power of $x_1$ works, since every element $M$ is a sum of such homogeneous pieces.

EDIT: This is also posted at SE, see here. Apparently the definition I used for the radical is not correct - instead it is the intersection of all the prime ideals containing the object. I would imagine that the above is not longer useful.

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Dear Andrew, Unless I am missing something, your definition of radical (the set of $x$ such that $x^n$ lies in the ideal $I$) coincides with the definition of radical as the intersection of prime ideals. So your proof looks reasonable to me. (For the second direction, it might be simpler to note that $M/N^*$ is graded, hence its annihilator is graded.) Regards, Matthew – Emerton Jun 7 '12 at 14:16
Thanks for clarifying Matthew. – Andrew Davies Jun 8 '12 at 8:06

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