Let $L=\{b^ma^n|m \space and \space n \space are \space coprime \}$ using Nerode theorem prove that $L$ is irregular.
From Nerode theorem I know that $L$ is regular if and only if the number of equivalence classes of $R_L$ (the relation defined in Nerode theorem) is finite, so I need to prove that there are infinite number of equivalence classes.
The first thing that came to mind from $L$'s definition is using Dirichlet's theorem, hence I tried:
Let $w_{m, i}=b^ma^i$, ($m,i$ are coprimes), and I prove that for $j\ne i$, ($m, j$ coprimes), $$w_{m, i} \not R_L w_{m, j}$$ Let $z=a^{m+ni}$, ($n$ an integer promised by Dirichlet's theorem such that $m+ni$ and $m$ are coprimes), so $$w_{m, i}z = b^ma^{m+ni+i}= b^ma^{m+(n+1)i}\in L$$ and $$w_{m, j}z = b^ma^{m+ni+j}\not\in L$$ But this isn't necessarily true as $m$ and $m+(n+1)i$ might not be coprimes and $m$ and $m+ni+j$ might be
I know from previous exercises that I need to find $w_i$ and show that for a word $z$ $$w_iz\in L \space and \space w_jz\not\in L \space\space (i\ne j)$$ and therefore there are infinite number of equivalence classe, but I find it difficult with all the prime/coprime
There is no need of advanced number theoretic argument to prove this result. I claim that if $p$ and $q$ are distinct primes, then $(b^p)^{-1}L \not= (b^q)^{-1}L$. Suppose by contradiction that $(b^p)^{-1}L = (b^q)^{-1}L$. Since $p$ and $q$ are coprime, one has $b^pa^q \in L$. It follows that $a^q \in(b^p)^{-1}L$, whence $a^q \in(b^q)^{-1}L$, that is, $b^qa^q \in L$, a contradiction.
It follows that the quotients $(b^p)^{-1}L$, for $p$ prime, are all distinct, and thus $L$ is not regular.