If we have an ideal of the form $I=\langle f(x), g(x)\rangle\subseteq\Bbb Z[x] $ should it stand that $I=\langle \gcd(f(x),g(x))\rangle$?
For example, if we have the ideal $I=\langle 2,x \rangle $ does it stand that $\gcd(2,x)=1$? Or does it not stand necessarily?
The ideal $(2,x)\subseteq\Bbb Z[x]$ is indeed $1$, since $1|2$ and $1|x$ and any polynomial dividing the irreducible $x$ would have to be a constant and a unit since it is monic. However $2\not |x$ so the gcd cannot be $2$. In particular, $(2,x)$ is not $(1)$ since $\Bbb Z[x]/(1)\cong 0$ but $\Bbb Z[x]/(2,x)=\Bbb F_2$ is the field with two elements.
Addendum
To see that $\Bbb Z[x]/(2,x)\cong\Bbb F_2$ we just use the definition of quotienting.
Let $t(x)\in\Bbb Z[x]$ then by definition if
$$t(x)=\sum_{n=0}^k t_nx^n$$
modulo $(2,x)$ we have that $x\equiv 0$ so $t(x)\equiv t_0\mod (2,x)$, also since $2\equiv 0$ we have that $t_0\equiv r\mod (2,x)$ with $r\in \{0,1\}$. So $\Bbb Z[x]/(2,x)\subseteq \Bbb F_2$ is a subring. We show it is non-trivial
Proof: Let $q(x),s(x)\in\Bbb Z[x]$ such that $xq(x)+2s(x)=1$, then since $q(0)=0$ it must be that the constant term of $s$ is ${1\over 2}$ which is obviously a contradiction, so no such polynomials exist, and $1$ is not a polynomial combination of $x$ and $2$ hence $1\not\in(2,x)$, and so it has a non-trivial image in $\Bbb F_2$ so that the map $\Bbb Z[x]/(2,x)\to \Bbb F_2$ is surjective.