$\phi$ in $O_K$ but not in $\mathbb{Z}[t]$ I have this problem:
Let $t$ be a root of the polynomial $f(x) = x³ + x² - 2x + 8$. Let $\phi = \displaystyle \frac{4}{t}$
and let $K = \mathbb{Q}(t)$.
I was able to show that $f(x)$ is irreducible, and that $\phi$ is in $O_K$. However, I am not sure how to show that $\phi$ is not in $\mathbb{Z}[t]$.
Any help is appreciated.
 A: Perhaps a little more simply than @awllower answers, I say
\begin{align*}
0&=t^3+t^2-2t+8\\
8/t&=-t^2-t+2\\
4/t&=1-\frac12t-\frac12t^2\,,
\end{align*}
and since $\{1,t,t^2\}$ is a basis for $K$ over $\mathbb Q$ (here’s where irreducibility gets used), the coefficients $1,-1/2,-1/2$ are uniquely determined.
A: Since you have shown that $f(x)$ is irreducible, if $t$ satisfies a polynomial of degree $3$, then that polynomial should be a multiple of $f(x)$. Now, if $p=at^2+bt+c$, then $at^3+bt^2+ct-4=0$, so that $a,b$ cannot be integers, i.e. $p$ cannot lie in $\mathbb Z[t]$.
P.S. In fact $a=b=-1/2$, $c=1$.
A: If $\rm\ \exists\,\phi,g\in\Bbb Z[t]\!:\,\ 4\, =\, t\, \phi(t) + f(t)\, g(t),\ $ eval at $\rm\ t=0\: \Rightarrow\: 4=8\,g(0),\ $ contra $\rm\ g(0)\in \Bbb Z\:$
Remark $\ $  So, more generally: $ $  if $\rm\,n\in\Bbb Z,\,\ f\in \Bbb Z[t]\,$ then $\rm\ t\mid n\,\ in\,\ \Bbb Z[t]/(f)\ \Rightarrow\ f(0)\mid n\,\ in\,\ \Bbb Z$
Therefore, defining $\rm\ norm(t)\,:=\, f(0),\ $ we have:  $\rm\ \ t\mid n\,\ in\,\ \Bbb Z[t]/(f)\iff norm(t)\mid n\,\ in\,\ \Bbb Z$
