Identifying quotient rings The question I have is from a past exam paper:
Identity if it is an integral domain, whether it admits a finite basis over a coefficient ring, or whether it is isomorphic to another ring.
The first one is
$$R = \mathbb{Z}[x]/(6x^2 -1, 2x-4)$$
My first angle of attack is to find if we can simplify the ideal $I=(6x^2-1,2x-4)$ in hopes of finding a monic irreducible polynomial by taking combinations of elements of the ideal to end up with $I = (f)$ to apply a substitution isomorphism, maybe this is too hopeful.
We start out by making a few computation on elements of the ring
\begin{align*}
6x^2-1 &= 0\\
2x*3x-1&=12x-1 = 0\\
\end{align*}
So we can use $12x-1=0$ and $2x-4$ to perhaps find if this ring is isomorphic to some modular ring.
$$6(2x-4) = 12x - 24 = 1-24 = 0 = 23 \in I$$
and
$$R = \mathbb{Z}[x]/(23,6x^2-1,2x-4) \cong \mathbb{Z}_{23}[x]/(6x^2-1,2x-4)$$
Next we can use this modulo $23$ in the following way to perform further computations:
$$4(6x^2-1) = 24x^2-4=x^2-4=0$$
We then can reduce the ideal to
$$I = (x^2-4,2x-4)$$ 
If $x^2 = 4$ and $2x=4$ we must have that $x = 2$ and $I = (x-2)$
We then write
$$R \cong \mathbb{Z}_{23}[x] /(x-2) \cong \mathbb{Z}_{23}[2] = \mathbb{Z}_{23}$$
Also, I know that this ring isn't an integral domain because we can take $23,1 \in R$ and get $23*1 = 0$, so therefore $23$ is a zero divisor.
Let me know if my working is correct, I feel a little bit iffy about reducing the ideal, thanks.
EDIT: A finite basis for $R$ would just be $\{a|a \in \mathbb{Z}\}$
 A: I juste realized how wrong. Is this better?
Let $f=6x^2-1$ and $g=2x-4$. The division algorithm gives $f=g(3x-6)+23$. 
We have $6x^2-1\equiv 23\mod g$ and $\mathbb{Z}[X]/(g)=\mathbb{Z}[x]/(x-2)\mathbb{Z}[x]\cong \mathbb{Z}$. Modding successively gives $\mathbb{Z}[x]\to \mathbb{Z}\to \mathbb{Z}/23\mathbb{Z}$.
A: You're mixing up things and arrive to a wrong conclusion.
Your strategy is good, it just needs some adjustment in wording.
Let $\xi$ denote the image of $x$ in the quotient ring; then, by assumption, $6\xi^2-1=0$ and $2\xi-4=0$. Therefore
$$
1=3\xi\cdot 2\xi=3\xi\cdot 4=6\cdot 2\xi=24
$$
In particular, $24-1=0$ in the quotient ring, which means $23\in I$.
The next step is noting $4\cdot 6\xi^2=4$, so $\xi^2=4$. Together with $2\xi=4$, this implies $\xi^2=2\xi$. Also
$$
1=6\xi^2=6\cdot2\xi=12\xi
$$
and we conclude that $2$ is invertible in the quotient ring. From $2\xi=4$, we conclude $\xi=2$.
So, we have proved that $23\in I$ and $x-2\in I$. Now we can do the final step: proving $I=(23,x-2)$. Clearly, $2x-4\in(23,x-2)$. Also
$$
6x^2-1=(6x+12)(x-2)+23\in(23,x-2)
$$
Thus your ring is
$$
\mathbb{Z}[x]/(23,x-2)\cong(\mathbb{Z}/23\mathbb{Z})[x]/(x-2)
\cong\mathbb{Z}/23\mathbb{Z}
$$
is a field.
