I'm stuck on an exercise (E.VIII.4.1) from Algebra by Hungerford. It is stated as follows (sorry for the bad title; I didn't know how to fit it):

Let $R$ be a commutative ring with identity and $I$ a finitely generated ideal of $R$. Let $C$ be a submodule of an $R$-module $A$. Assume that for each $r\in I$ there exists a positive integer $m$ (depending on $r$) such that $r^mA\subseteq C$. Show that for some integer $n$, $I^nA\subseteq C$.

Along with the exercise he provides a hint to reference two previous theorems:

If $R$ is a ring, then $$ \left(\sum_{i=1}^{n}a_i\right)\left(\sum_{j=1}^{m}b_j\right)=\sum_{i=1}^n\sum_{j=1}^ma_ib_j $$ for all $a_i,b_j\in R$.


Let $R$ be a ring and $X\subseteq R$. If $R$ has an identity and $X$ is in the center of $R$, then the ideal $(X)$ consists of all finite sums $r_1a_2+\cdots +r_na_n$ ($n\in\mathbb{N}^*; r_i\in R;a_i\in X$).

I imagine that the proof goes something like: If $X=\{x_i\in I\,|\,i\in J\}$ is the generating set of $I$, then there exist positive integers $m_i$ such that $x_i^{m_i}A\subseteq C$ for each $x_i$, and then since $X$ is finite, you pick the largest one (or possibly a product/sum) and then show that works for all $r\in I$. The only problem is I'm having trouble dealing with elements of the form $$ \left(r_1x_1+\cdots +r_nx_n\right)^m $$ Would some sort of induction-type argument (possible on the number of generators) work? Or am I going about this the wrong way? Thanks in advance.


Your idea is completely correct and, in fact, I think you got pretty close to a nice solution: take for example

$$\;m=\sum_{i\in J}m_i\implies$$

using MT , you now get

$$\left(\sum_{i\in J}r_ix_i\right)^m=\sum_{k_1+\ldots+k_n=m}\binom m{k_1,...,k_n}\prod_{1\le t\le n} (r_tx_t)^t$$

and multiplication by $\;A\;$ gets us in $\;C\;$ .

Perhaps some induction could help to make things clearer, with $\;n=2\;$ applying the usual Newton's Binomial Theorem and etc.

  • $\begingroup$ Thank you so much! In the back of my head I was thinking about the multinomial theorem, but I was distracted trying to apply that first hint directly. $\endgroup$ – Blake Apr 11 '16 at 10:36

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