Show that if $R/A$ is IBN then $R$ is IBN.

Question

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.

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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.

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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!

Answer 1

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.