TITLE: What is $\mathbb{Z}[x]/(6,2x-1)$ isomorphic to?
QUESTION [6 upvotes]: What is $\mathbb{Z}[x]/(6,2x-1)$ isomorphic? Since $\mathbb{Z}[x]/(6,2x-1)$ is isomorphic to $\frac{\mathbb{Z}[x]/(6)}{(2x-1)}$ then it's isomorphic to $\mathbb{F}_6[\frac{1}{2}]$. But on the other hand, the ideal $(6,2x-1)$ is equal to $(3,x-1)$ then $\mathbb{Z}[x]/(6,2x-1)$ is isomorphic to $\mathbb{F}_3/(x-1)$ which is isomorphic to $\mathbb{F}_3$? 
However, $\mathbb{F}_6[\frac{1}{2}]$ is not isomorphic to $\mathbb{F}_3$, so where am I wrong?

REPLY [4 votes]: You can use, o prove, the following:
Claim: An ideal in $\;\Bbb Z[x]\;$ is maximal iff it is of the form $\;\langle\,p,\,f(x)\,\rangle\;$ , with $\;p\;$ a prime number and $\;f(x)\in\Bbb Z[x]\;$ an element such that $\;f(x)\pmod p\;$ is irreducible in $\;\Bbb F_p[x]\;$ .
We're then almost done when we realize that $\;\langle\,6,\,2x-1\,\rangle=\langle\,3,\,2x-1\,\rangle\;$ and $\;2x-1\in\Bbb F_3[x]\;$ is irreducible and thus the quotient ring is a field...
Finally, the steps from $\;\Bbb Z[x]/\langle6,\,2x-1\rangle\;$ to (isomorphism) $\;\Bbb F_3\;$ are very similar to what was done in the other answer: using the second-third isomorphism theorems and etc.