$\mathbb{Z}_6[x]$/$\langle2x-3\rangle$ isomorphic to what?

Question

To what common ring is $\mathbb{Z}_6[x]$/$\langle2x-3\rangle$ is isomorphic?

Obviously not $\mathbb{Z}_6[\frac{3}{2}]$... Thanks!

Answer 1

Hint $\ {\rm mod}\ 6\!:\ (2x\!-\!3)(3x\!+\!2) \equiv x\,\Rightarrow \langle 2x\!-\!3\rangle\equiv \langle x,3\rangle\Rightarrow\,\Bbb Z_6[x]/\langle 2x\!-\!3\rangle=\Bbb Z_6[x]/\langle x,3\rangle\cong\ldots$

Answer 2

By the Chinese remainder theorem $\def\Z{\Bbb Z}\Z_6\cong\Z_2\times\Z_3$, and using the projections on the two factors your quotient becomes a product of quotients: $\Z_6[x]/\langle2x-3\rangle\cong\big(\Z_2[x]/\langle1\rangle\big)\times\big(\Z_3[x]/\langle-x\rangle\big)$. The first factor is the trivial ring and the second factor is just $\Z_3$. This uses $(R\times S)[x]\cong R[x]\times S[x]$, which is a generally valid identification via the obvious projections.

It might be good to give a bit more detail on how this argument works. Reducing modulo $2~$and$~3$ we find $2x-3\equiv1\pmod2$ respectively $2x-3\equiv-x\pmod2$, which one can write combined as $2x-3\equiv(1,-x)\pmod{2,3}$. But since the elements $1$ of the factors $\Z_2$ and $\Z_3$ respectively lift to $3\equiv(1,0)\pmod{2,3}$ and to $4\equiv(0,1)\pmod{2,3}$ in$~\Z_6$, these projections of $2x-3$ can also be represented by genuine multiples of it in $\Z_6[x]$, namely by $3\times(2x-3)=3$ and $4\times(2x-3)=2x$. Using multiples of the first of those one can see that any polynomial in $\Z_6[x]$ is congruent modulo $\langle2x-3\rangle$ to its projection to $\Z_3[x]$ inside $\Z_6[x]$ (this reflects the first factor being the trivial ring), and using the second of those that projection can be further reduced to its constant term (which is one of $0,2,4$ since elements of $\Z_3[x]\subset\Z_6[x]$ have even coefficients, but taking them modulo $3$ gives the more common representation as element of $\Z_3$).