Trying to determine $\mathbb{Z}[x] / (6, 2x - 1)$ 
4.3. Identify the following rings:

*

*$\mathbb{Z}[x] / (x^2 - 3, 2x + 4)$,


*$\mathbb{Z}[i] / (2 + i)$,


*$\mathbb{Z}[x] / (6, 2x - 1)$,


*$\mathbb{Z}[x] / (2 x^2 - 4, 4 x - 5)$,


*$\mathbb{Z}[x] / (x^2 + 3, 5)$.
(Original image)

I'm trying to solve (3). So, I took the canonical homomorphism $\phi \colon \mathbb{Z}[x] \to \mathbb{Z}_6[x]$ as
$$
   \phi(a_n x^n + \dotsb + a_0)
   = (a_n \bmod 6) x^n + \dotsb + (a_0 \bmod 6) \,.
$$
(Here, $\mathbb{Z}_6 = \mathbb{Z} / 6\mathbb{Z}$).
Then, the kernel of this surjective homomorphism is $(6)$ (the ideal generated by $6$). So, $\mathbb{Z}[x]/(6)$ is isomorphic to $\mathbb{Z}_6[x]$ (by the first isomorphism theorem). Now, to find $\mathbb{Z}[x]/(6, 2x-1)$, I will need to find
$$
  \mathbb{Z}_6[x]/((2 \bmod 6) x + (-1 \bmod 6))
  = \mathbb{Z}_6[x]/(2x + 5) \,.
$$
I have seen somewhere that the answer is $\mathbb{Z}_3$, but how should I analyze this further? $2$ is also not a unit in $\mathbb{Z}_6$, so that I could apply division rule and simplify this into a finite set.
 A: Good start! I'll first give a high-level, "intuitive" explanation of what's going on, then sketch a rigorous proof.

High Level Intuition $\mathbb{Z}_6[x]/(2x-1)$ can be thought of as $\mathbb{Z}_6[\frac{1}{2}]$, because we've adjoined an element ($x$) such that $2$ times $x$ is equal to $1$. Also, by the chinese remainder theorem, $\mathbb{Z}_6 \cong \mathbb{Z}_2 \times \mathbb{Z}_3$. So, we have $\mathbb{Z}_6[x]/(2x-1) \cong (\mathbb{Z}_2 \times \mathbb{Z}_3)[\frac{1}{2}]$.
When we try to adjoin $1/2$ to $\mathbb{Z}_2$, though, something goes wrong. We get
$$1 = 2 \cdot \frac{1}{2} = 0 \cdot \frac{1}{2} =0,$$
forcing the ring to be zero!
On the other hand, $2$ is already a unit in $\mathbb{Z}_3$, so $\mathbb{Z}_3[\frac{1}{2}]$ is just $\mathbb{Z}_3$.
Overall, this means
$$\mathbb{Z}_6[x]/(2x-1) \cong (\mathbb{Z}_2 \times \mathbb{Z}_3)[\frac{1}{2}] \cong \mathbb{Z}_2[\frac{1}{2}] \times \mathbb{Z}_3[\frac{1}{2}] \cong 0 \times \mathbb{Z}_3 \cong \mathbb{Z}_3.$$
With some tools, ideas, and language you'll probably learn later, this can be made into a rigorous proof! But I would guess you haven't learned about localization yet, so we should try for a more hands-on, direct proof.

Proof Sketch To show that $\mathbb{Z}_6[x]/(2x-1) \cong \mathbb{Z}_3$, we simply need to construct an isomorphism between the two rings. By the above intuition, we should try to send $x$ to the multiplicative inverse of $2$ in $\mathbb{Z}_3$, which is $2$.
So, let's do it! Let $\pi : \mathbb{Z}_6 \to \mathbb{Z}_3$ be the canonical projection (reduction modulo $3$). Then define $\varphi : \mathbb{Z}_6[x] \to \mathbb{Z}_3$ by $\varphi(f) = \pi(f(2))$. That is, we map $x \mapsto 2$ and then we reduce modulo $3$. Next, show that $\ker \varphi = (2x-1)$. By the first isomorphism theorem, this will complete the proof!
A: The ideal $I = (6, 2x - 1)$ contains the element $3(2x - 1) = 6x - 3$, but also the element $6x$, and therefore the element $3 = 6x - (6x - 3)$.
But $6$ is a multiple of $3$, so
$$
  I = (6, 2x - 1) = (6, 2x - 1, 3) = (3, 2x - 1) \,.
$$
Replacing $6$ by $3$ in your approach, we find that
$$
  ℤ[x] / I = ℤ[x] / (3, 2x - 1) ≅ ℤ_3[x] / (2x - 1) \,.
$$
In $ℤ_3[x]$ we have $2x - 1 = -x - 1 = -(x + 1)$ and therefore
$$
  ℤ_3[x] / (2x - 1) = ℤ_3[x] / (x + 1) ≅ ℤ_3 \,.
$$
PS.
We could also further simply the ideal $I$, by replacing the generator $2x - 1$ with $(2x - 1) - 3x = -x-1 = -(x + 1)$.
Then
$$
  I = (3, 2x - 1) = (3, -(x + 1)) = (3, x + 1) \,.
$$
In this form we can then mod out the two generators $3$ and $x + 1$ in either order.
