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

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!

• – user26857 Apr 1 '20 at 13:58

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.