Why $t-7$ and $4$ are not coprime in $\mathbb Z[t]$? I have to compute $(t-7,4)/(t-7)$ in $\mathbb Z[t]$. What I did is : since $t-7$ and $4$ are coprime, $(t-7,4)=\mathbb Z[t]$. Therefore, $$(t-7,4)/(t-7)\cong \mathbb Z[t]/(t-7)\cong \mathbb Z,$$
since $t-7$ is irreducible in $\mathbb Z[t]$. The Teacher assistant gave me a mark of $0$ out of $10$ and said that $(t-7,4)/(t-7)\cong 4\mathbb Z$. So, why $t-7$ and $4$ are not coprime, i.e. what would be the gcd ? $\color{red}{(solved)}$ And how to prove that $(t-7,4)/(t-7)\cong 4\mathbb Z$ ? this doesn't natural for me.
 A: $\gcd(t-7,4)=1$ and $(t-7,4)\ne\mathbb Z[t]$.
However $(t-7,4)/(t-7)\not\simeq\mathbb Z/4\mathbb Z$ (as groups, as rings, or whatever you want!) since the quotient is infinite. Instead it is an ideal of $$\mathbb Z[t]/(t-7)\simeq\mathbb Z\tag{*}.$$
Furthermore, the quotient $(t-7,4)/(t-7)$ is an ideal of $\mathbb Z[t]/(t-7)$ while $4\mathbb Z$ is an ideal of $\mathbb Z$. As I have already mentioned the rings are isomorphic, and the isomorphism is pretty obvious: define $\mathbb Z[t]\to\mathbb Z$ by $t\mapsto 7$. (Notice that the kernel is $(t-7)$.) Then $(t-7,4)/(t-7)$ corresponds to $4\mathbb Z$.
A: Your problem is that you've confused two notions: a GCD domain and a principal ideal domain. The GCD of two elements $a,b\in R$ with $R$ a ring may be defined as the smallest principal ideal $(c)$ containing $(a,b)$. In your case, the GCD of $4$ and $t-7$ is indeed $(1)$. But $\mathbb Z[t]$ is not a principal ideal domain: not every ideal is principal. Thus even though $(1)$ is the smallest principal ideal containing $(4,t-7)$, we cannot conclude that $(1)=(4,t-7)$, as indeed it does not.
