$R^\mathbb{N}$ is not finitely generated as an $R$-module

Question

There is this exercise in my course note in which I have to prove following statements :

Assume $R$ is commutative and $1\neq 0$.

$(i)$ $R^\mathbb{N}$ is not finitely generated as an $R$-module

$(ii)$ $R^\mathbb{N}$ is finitely generated as an $R^\mathbb{N}$-module

The seconde statement $(ii)$ seemed obvious to me as every element in $R^\mathbb{N}$ can be obtained directly from $R^\mathbb{N}$ by multiplication with the multiplicative identity $1_{R^\mathbb{N}}$.

But the first statement $(i)$, I was not so sure about. By my understanding $R^\mathbb{N}$ is the set of $\{R_1,R_2,..\}$ Where every $R_i = R$ and $i \in \mathbb{N}$ and an element of $R^\mathbb{N}$ would be an $n$-tuple with elements of $R$ and $n \in \mathbb{N}$.

So for the proof I first assumed there was a finite generating set $S$ for the $R$-module $R^\mathbb{N}$. Since $S$ is finite there is always an element in $S$ that is an $n$-tuple, with the greatest $n$, that is there is no $(n + 1)$-tuple in $S$ and hence any $(n+1)$-tuple can't be obtained from $R$ and $S$.

Now to me the way an element in $R^\mathbb{N}$ is defined is vague as well as the real meaning of $\mathbb{N}$ in the exponent. I know $\mathbb{N}$ is used as the indexing set, but does that mean every natural number can be assigned to an element in $R^\mathbb{N}$ ? Also I never used the fact that $R$ is commutative and $1 \neq 0$...

Thank you for reading my question.

$R^{\Bbb N}$ is the set of sequences $(a_1,a_2,\ldots)$ with each $a_i\in R$. — Angina Seng, Apr 28 '20 at 15:42
@AnginaSeng So for example if $R$ = $\mathbb{Z}$ then for example (-2,4,5) is an element of $R^\mathbb{N}$ , as well as (1,1,0,6)? — Moeee, Apr 28 '20 at 15:48
Okay right, so $R^\mathbb{N}$ has infinite elements in the tuple — Moeee, Apr 28 '20 at 15:56

score 1 · Accepted Answer · answered Apr 28 '20 at 15:41

1

I'll give a hint: can you prove that if $R^{\textbf N}$ is finitely generated, then so is $R^n$ for any $n\in\Bbb N$. Can you also prove that $R^n$ requires at least $n$ generators?

answered Apr 28 '20 at 15:41

Angina Seng

158,341

Well the fact that $R^n$ requires at least $n$ generators reminds me of how a basis for a vector space $\mathbb{R}^n$ has at least n elements. – Moeee Apr 28 '20 at 15:54
1

The basic idea is the same. Look into "free modules'. – Aniruddh Agarwal Apr 29 '20 at 03:15

$R^\mathbb{N}$ is not finitely generated as an $R$-module

1 Answers1

Linked