1
$\begingroup$

Let $R$ be a ring and $A$ is an ideal of $R$. Show that if $R/A$ is IBN then $R$ is IBN.

As a hint to this proof it was given that $W/AW$ is a $R/A$-module.

The idea: Using what was given, if I'm able to show that if $W$ (which is itself a $R$-module) has a basis of cardinality $n$ then $W/AW$ will also have a basis of the same cardinality we'll be done. Why? Suppose that we've proved what I said and consider two basis for $W$ with different cardinality. From that we'll be able to create two basis for $W/AW$ with different cardinalities, but that's a contradiction since $W/AW$ is a $R/A$-module and $R/A$ is IBN.


The proof: Let's prove that if $W$ has a basis $B$ such that $|B| = n$ then $W/AW$ will have a basis $B'$ such that $|B'|=n$.

Suppose that $\{w_1,\cdots,w_n\}$ is basis for $W$. That implies that $W = Rw_1 \oplus \cdots \oplus Rw_n$. Now consider an element of $w+AW \in W/AW$. It follows: \begin{align*} w+AW &= \big(r_1w_1 + \cdots + r_nw_n\big) + AW\\ &= (r_1+A)(w_1+AW) + \cdots + (r_n+A)(w_n+AW)\\ &\implies W/AW = (w_1+AW) + \cdots + (w_n+AW) \end{align*} And it remains to show that this sum is actually a direct sum. For that, suppose we have two different compositions of a $w+AW \in W/AW$: \begin{align*} (r_1'+A)(w_1+AW) + \cdots + (r_n'+A)(w_n+AW) &= (r_1+A)(w_1+AW) + \cdots + (r_n+A)(w_n+AW)\\ \big(r_1w_1 + \cdots + r_nw_n\big) + AW &= \big(r_1'w_1 + \cdots + r_n'w_n\big) + AW \end{align*} And from that it follows that: \begin{align*} r_1w_1 + \cdots + r_nw_n - (r_1'w_1 + \cdots + r_n'w_n) &= 0\\ (r_1-r_1')w_1 + \cdots + (r_n-r_n')w_n &=0 \end{align*} and since $W = Rw_1 \oplus \cdots \oplus Rw_n$ we have that $r_i = r_i'$ for $i \in \{1,\cdots,n\}$. Therefore $\{w_1 + AW, \cdots, w_n + AW\}$ is a base for $W/AW$ of the same cardinality of the base of $W$ that we initially had.


What do you think about the strategy that I used? Is it correct for you? What would you do in order to prove the first statement without the hint that was given?

Just a side note, I'm new to abstract algebra and ring/module theory. So if you want to provide a proof as an answer to my previous question, please, if it's possible, don't use advanced results in the theory.

Thanks!

$\endgroup$
1
1
$\begingroup$

The easy way to prove this is to note that the definition of IBN for a ring $S$ amounts to "if there exist a matrix $A$ is an $n\times m$ and $B$ is an $m\times n$ matrix over $S$ such that $AB=I_n$ and $BA=I_m$ then $m=n$."

For if you had an isomorphism of $S^n\to S^m$, after selecting a basis, you would have exactly two such matrices: one for the isomorphism and one for its inverse.

Now if you supposed you had $A$ and $B$ over $R$, notice that if you apply the quotient map $R\to R/A$ you would get two matrices $A'$ and $B'$ over $R/A$ which satisfy $A'B'=I_n$ and $B'A'=I_m$, which implies $m=n$ since $R/A$ has the IBN. It follows that $R$ has the IBN property as well.

$\endgroup$
3
  • $\begingroup$ I'll think about that! By the way, my proof is indeed correct? Thanks! $\endgroup$ – Bruno Reis Nov 6 '19 at 20:49
  • $\begingroup$ @BrunoReis It looks like it's using a similar idea, but unfortunately I do not have time to read it in detail now. $\endgroup$ – rschwieb Nov 6 '19 at 20:51
  • $\begingroup$ No problem mate. If you can when you have time I'll appreciate it $\endgroup$ – Bruno Reis Nov 6 '19 at 20:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.