understanding finitely generated ideals

Question

Let $R$ be a commutative ring and $A \subset R$ be any ideal. The ideal $A$ is finitely generated if every $a \in A$ can be written in the form $a = r_1a_1+r_2a_2+ \cdot + r_na_n$ where $r_i \in R$ and we write $A = \langle a_1, a_2, \cdots , a_n\rangle$.

Is it correct to conclude that any element of the form $r_i^ma_i^n \in A?$

My reasoning: Since any $a\in A$ can be written as $ a = r_1a_1+r_2a_2+ \cdots + r_na_n$,

we get $a_i \in A \implies a_i^2 \in A \implies a_i^n \in A \implies r_ia_i^n \in A \implies r_i^ma_i^n \in A$

Answer 1

Your definition is not written correctly. In your definition, the elements $a_1,\ldots,a_n$ are not specified and so are free in your sentence. That means that either you are saying

For all $a_1,\ldots,a_n$ (where?) every $a\in A$ can be written as ...

or else you allow the $a_i$ to depend on $A$. In the first case, very few ideals would meet the condition; in the latter, all ideals would meet the condition.

The correct way to phrase it is to specify the $a_i$ first. So it should read:

An ideal $A$ of $R$ is finitely generated if there exist (a finite collection of) elements $a_1,\ldots,a_n\in A$ such that every $a\in A$ can be written in the form $a=r_1a_1+\cdots + r_na_n$, with $r_i\in R$. If that is the case, we write $A=\langle a_1,\ldots,a_n\rangle$.

But note that what you are asking about has nothing to do with whether the ideal is finitely generated. Simply:

If $A$ is an ideal, and $a\in A$, then for all $r\in R$ and all $n\geq 1$, $m\geq 1$, $ra^n\in A$.

If $n=1$, this follows because $A$ is an ideal, so $a\in A$, $r^m\in R$ implies $r^ma\in A$. For $n\gt 1$, you can let $s=r^ma^{n-1}\in R$, and then $a\in A$, $s\in R$ implies $sa = r^ma^{n-1}a = r^ma^n\in A$.

Nothing to do with finite generation, everything to do with being a ring and an ideal.

Answer 2

Is it correct to conclude that any element of the form $r_i^ma_i^n \in A?$

Yes.

Here is a simpler way. Take $r = r_i^m a_i^{n-1}$, then by definition of ideal $ra_i \in A,$ so $r_i^ma_i^n \in A.$