I'm taking an abstract algebra class and we are introducing quotient rings specifically polynomial quotient rings and I'm trying to work out some example problems, but I cannot figure out a general way to approach problems where we're asked to identify something like $\Bbb Z[x]/I$.
The example I have is that we can identify $\Bbb Z[i]/(i-2)$ to $\Bbb F_5$ first by killing $g=x-2$ and then killing $f=x^2+1$.
So with this approach if I have for example $$\Bbb Z[x]/ (6, 2x-1)$$ then if I first kill $2x-1$ and then $6$ I should be able to arrive to some isomorphic ring to this?
Killing $2x-1$ is apparrently done by defining a map $\Bbb Z[x] \to \Bbb Z[1/2]$ where $x \mapsto 1/2$ and since the kernel of this map is generated by $(2x-1)$ I'll get an isomorphism $$\Bbb Z[x]/(2x-1) \cong \Bbb Z[1/2].$$
Now following this I should kill $6$ in $\Bbb Z[1/2]$ to get some kind of isomorphism from $\Bbb Z[1/2]$ to some other ring and I suppose then I can conclude that $\Bbb Z[x]/ (6, 2x-1)$ is isomorphic to the obtained ring?
How do I go about killing $6$ in $\Bbb Z[1/2]$? I don't think I understand how to construct something like this and what should the target ring even be in this case?
Solution 1. As you were told, $\mathbb Z[x]/(6,2x−1)\simeq\mathbb Z_6[x]/(2x-1)$. But $$\mathbb Z_6[x]/(2x-1)\simeq \mathbb Z_2[x]/(2x-1)\times\mathbb Z_3[x]/(2x-1)\simeq\mathbb Z_3.$$
Solution 2. You can use the rings of fractions: $$\mathbb Z_6[x]/(2x-1)\simeq S^{-1}\mathbb Z_6,$$ where $S=\{\bar 1,\bar 2,\bar 4\}$, and then notice that $S^{-1}\mathbb Z_6\simeq\mathbb Z_3$.
Solution 3. As you noticed, $$\mathbb Z[x]/(2x-1)\simeq\mathbb Z[\frac12].$$ Then $$\mathbb Z[x]/(6,2x-1)\simeq\mathbb Z[\frac12]/6\mathbb Z[\frac12].$$ But $6\mathbb Z[\frac12]=3\mathbb Z[\frac12]$, and you arrived at $\mathbb Z[\frac12]/3\mathbb Z[\frac12]$. This is isomorphic to $S^{-1}\mathbb Z_3$, where $S=\{\bar 1,\bar 2\}$. Since $\bar 2$ is invertible in $\mathbb Z_3$, you actually got $\mathbb Z_3$.