I am given $f(x)=x^4+x^3+x^2+x+\overline{1}\in\mathbb{Z}/2[x]$ and I have shown that this is irreducible.
I have the map $φ:\mathbb{Z}/2[x]\rightarrow \mathbb{Z}/2[x]/(f(x))$ given by $φ(g(x)) = \overline{g(x)}$. I have shown that it is a ring homomorphism. I am looking to show that $φ$ induces an isomorphism $\overline{φ}:\mathbb{Z}/2[x]/(f(x))\rightarrow\mathbb{Z}/2[x]/(f(x))$ which is different from the identity.
To do this I can show that $\ker(φ)=(f(x))$ and use the First Isomorphism Theorem. Of course,
$φ(g(x))=0 \iff g(x).g(x) + (f(x)) = (f(x))$,
but I don't know where to go from here. I know I need to show that the above is true if and only if $g(x) + (f(x)) = (f(x))$. If the product of $2$ elements $ab$ belongs to an ideal does that mean $a,b$ both belong to that ideal? I don't suppose this is true in general, so I think this has something to do with $f$ being irreducible.
I am also confused as to what the question means when it says 'different from the identity'. Could someone also explain what this part means?
$ab \in I$ rarely implies $a,b\in I$. A very important kind of ideal are prime ideals satisfying $ab\in I \Rightarrow a\in I \lor b \in I$. For our purpose this property suffices.
As $\Bbb Z/2\Bbb Z$ is a field, $\Bbb Z/2\Bbb Z[X]$ is a principal ideal domain and irreducible elements are prime elements. Hence the ideal $(f)$ is a prime ideal. This means that $\phi(g) = 0 \Leftrightarrow gg \in (f) \Leftrightarrow g \in (f)$ (prime yields $\Rightarrow$, the converse always holds).
For the other question: It is important to make a difference between isomorphism and identity. You are asked to show that there is some element in $\Bbb Z/2\Bbb Z[X]/(f)$, which is not mapped to itself. For example $\Bbb Z \rightarrow \Bbb Z, k \mapsto -k$ is an isomorphism of groups, but not the identity.