I am reading T.S.Blyth's Module Theory where it is written that two bases of an $R-$ module may have different cardinalities but if the ring $R$ is commutative then any two bases(if exist) of an $R$ module $M$ are equipotent.
In the proof it uses a fact that if a ring is commutative then it has a maximal ideal and then the proof follows.
But I have proved that any ring with identity has a maximal ideal then why is commutative needed here??
Any help how to sort this out?
Let us say that your $M$ is a left $R$-module.
If you try to reuse the same proof in the noncommutative case, you will fail: here is why.
The concept of maximal ideal can be generalized in two reasonable ways: you can try to use a left maximal ideal or a two-sided maximal ideal (see Wikipedia). (We exclude immediately the third possibility of using a right maximal ideal $m$, since here we would not even know how to build the quotient $M/mM$ as an $R$-module).
If you use a maximal left ideal $m\subset R$, you have a well-defined multiplication $R/m\times M/mM\to M/mM$, but $R/m$ might be no longer a ring (let alone a field): to guarantee this you need that $m$ is a two-sided ideal, which is not true in general!
If instead you try to use a maximal two-sided ideal, the quotient could well be a noncommutative ring: this happens for instance if $R=S\times S$, where $S$ is a simple ring, and $m=R\times\{0\}$.
The proof makes substantial use of the fact that the ring $R$ is commutative, so that the quotient $R/\mathfrak{m}$ modulo a maximal ideal $\mathfrak{m}$ is a field.
What you can say in the noncommutative case is that $R/\mathfrak{m}$ is a simple ring (no nonzero proper ideals), when $\mathfrak{m}$ is a maximal (two-sided) ideal.
Alas, there exist simple rings which don't have “invariant basis number”. An easy example is given by the ring $R$ of endomorphisms of the vector space $V$ over a field, assuming that $V$ has an infinite basis.
No endomorphism ring of a non finitely generated vector space has invariant basis number, because $R\oplus R\cong R$ (it's a nice exercise based on the fact that an infinite basis can be partitioned into to disjoint subsets with the same cardinality).
If $\mathfrak{m}$ is a maximal two-sided ideal of $R$, then $R/\mathfrak{m}$ doesn't have invariant basis number as well as $R$. Indeed, if $M$ is a finitely generated free left $R$-module, then $M/\mathfrak{m}M$ is a finitely generated free left $R/\mathfrak{m}$-module. If $R/\mathfrak{m}$ had invariant basis number, $R$ would as well.
