Quotient rings and identification I am trying to identify the following quotient ring using Adding relation method from Artin.
The ring is $\mathbb{Z}[x]/(6,2x-1)$. I did find solution to a similar problem on MSE. Hence I tried to follow the method. Here is my solution:
$2(2x-1)+4=4x-4$. So we get $\mathbb{Z}[x]/(6,4x-4)$ So, lets first kill 6. The ideal (6) is kernel for the homomorphism $\mathbb{Z}[x] \rightarrow \mathbb{Z_6}$ So, the residue of x is 6. Hence $4x-4=4(6)-4=18$. Therefore the ring is the one which contains 18 elements? Is this correct approach?
 A: Hint $\,\ 6 \equiv 0\ $ times $\ 1/2 \equiv x\,\Rightarrow\, 3\equiv 0,\ $ so $\ (6,2x\!-\!1) = (3,2x\!-\!1) = (3,2\!-\!x).\ $   
Therefore $\ \Bbb Z[x]/(6,2x\!-\!1)\, \cong\, \Bbb Z[x]/(3,2\!-\!x)\,\cong\, \Bbb Z_3[x]/(2\!-\!x)\,\cong\, \Bbb Z_3$
A: Consider the simplification of the ideal $(6,2x-1) = I$:
We find that $3(2x-1) \in I$ since ideals are closed under multiplication. Therefore we get
\begin{align*}
6x - 3 &= 0\\
-3 &= 0\\
(-1)(-3) &= 0\\
3 &=0
\end{align*}
Therefore we have that $3 = 0$ in the ideal $I$.
Note also that $(6,2x-1) = (3,2x-1)$, (we can check this by showing that the generators give the same elements in both ideals).
Next we want to simplify the polynomial term so we can use a substitution map, since $1/2 \notin \mathbb{Z}$, we can $not$ substitute $x \rightarrow 1/2$. But we can perform the following trick:
Since $3=0$, we can write
\begin{align*}
2x-1 &= 0\\
2x-1 &= 3\\
2x-4&=0\\
2(x-2)&=0\\
\end{align*}
Therefore $x-2 = 0$ in our new ideal, and in fact
$(6,2x-1) = (3,2x-1) = (3,x-2)$, which you can again, verify by showing that the elements generated by the generating elements give the same ideal. The last part is simple
$$\frac{\mathbb{Z}[x]}{(6,2x-1)} =\frac{\mathbb{Z}[x]}{(3,x-2)}$$
Since the ideals are the same.
By the substitution map $3 \rightsquigarrow 0$, we have
$$\frac{\mathbb{Z}[x]}{(3,x-2)} \cong \frac{\mathbb{Z}_3[x]}{(x-2)}$$
By the substitution map $x \rightsquigarrow 2$, we have
$$\frac{\mathbb{Z}_3[x]}{(x-2)} \cong \mathbb{Z}_3[2] = \mathbb{Z}_3$$
Since $2 \in \mathbb{Z}_3$.
A: $2(2x-1)+4=4x-4$ in $A=\mathbb{Z}[x]/(6,2x-1)$, it's true, but neither of these elements is zero. Rather $2(2x-1)+4=4$ since $2x-1=0$. Also $(6)$ is the kernel of $\mathbb{Z}[x]\to\mathbb{Z}_6[x]$ rather than $\mathbb{Z}_6$; that might have been just a typo on your part. Let's start over. 
By the Third Isomorphism Theorem $A$ is just $\mathbb{Z}_6[x]/(2x-1)$. Now this is a strange relation to have in a ring of characteristic $6$, because it seems to make $2$ both a unit and a zero divisor: we have $2x=1$ and $2*3=0$. To make these properties consistent we must in fact also have $3=0$, i.e. $A\simeq \mathbb{Z}_3[x]/(2x-1)$. (We can find the relation $3=0$ more explicitly: $3=6x=0$.) There's one last step in the identification, which I'll leave to you (hint: there's already an element in $\mathbb{Z}_3$ which is half of $1.$)
