Elementary way to show $\mathbb{Z}[e^{2 \pi i / 5}]$ and $\mathbb{Z}[X] / (X^4 + X^3 + X^2 + X + 1)$ are isomorphic I'd like to show that $\mathbb{Z}[e^{2 \pi i / 5}]$ is isomorphic to $\mathbb{Z}[X] / (X^4 + X^3 + X^2 + X + 1)$.
I'm thinking about applying the isomorphism theorem to
$f:\mathbb{Z}[X] \to \mathbb{Z}[e^{2\pi i / 5}], X \mapsto e^{2\pi i / 5}$.
Showing that $\operatorname{ker}(f) \supset (X^4 + X^3 + X^2 + X + 1)$ is straightforward:
$\operatorname{ker}(f)$ is prime, and $X^5 - 1 = (X - 1)(X^4 + X^3 + X^2 + X + 1) \in \operatorname{ker}(f)$, but $X - 1 \notin \operatorname{ker}(f)$, thus $X^4 + X^3 + X^2 + X + 1 \in \operatorname{ker}(f)$. Alternatively, and even more elementary, plugging in the value verifies that it is in the kernel.
Is there a quick way to show the reverse? Or is there another, better way to show it?
(I can't comment, so I have to answer to comments in this way: Showing irreducibility by comparing coefficients is possible, but how would it help?)
 A: $$f:\Bbb{Z}[X]\to\Bbb{Z}\left[{e^{2\pi i\over 5}}\right]\\X\to e^{2\pi i\over 5}$$
$f$ is ring homomorphism,
$$\Bbb{Z}[X]/\langle p(x)\rangle\simeq \Bbb{Z}\left[e^{2\pi i/5}\right]$$ where $p(x)$ monic irreducible polynomial and have root $e^{2\pi i/5}$ 
we have $p(x)=x^4+x^3+x^3+x^2+x+1$
A: $\newcommand{\Z}{\mathbb{Z}}$
$\newcommand{\Q}{\mathbb{Q}}$
$\newcommand{\cont}{\operatorname{cont}}$
$\newcommand{\Ker}{\operatorname{Ker}}$
Let $\phi(x)=x^4+..1$ and let $\Ker$ be the kernel of $\Z[x] \xrightarrow{evaluation} \Z[e^{2 \pi i/4}]$
Fact(see my comment above): The substitution map induces an iso $\Q[x]/(\phi(x)) \cong \Q[e^{2\pi i/5}]$.
Lemma: If $f \in \Ker$,$f$ has content 1, $\Rightarrow$ $f$ is in $(\phi) \in \Z[x]$.
Proof:
Now let $f \in \Z[x]$ satisfy $f(e^{2 \pi i/5})=0$ and also be of content 1.  Then $f =\phi h$ for some $h \in \Q[x]$.   Write $c h(x)=H(x)$ where $H(x)$ has content 1, and $c \in \Z$. Then $c=c \cont(f)=\cont(\phi)\cont(H)=1\cdot 1$. QED.
Now let $f \in Ker$ be arbitrary.  Then $f=\cont(f)  \cdot F$ where $F$ has content $1$.  $F$ is in $(\phi) \subset \Z[x]$ by the lemma, so $f$ is too.
