To show $\mathbb Z[t^4,t^7]$ is not integrally closed in $\mathbb Z[t]$ is it fine to say that $t$ is a root of $x^4-t^4$ which is a monic polynomial with coefficients in $\mathbb Z[t^4,t^7]$, but $t\notin \mathbb Z[t^4,t^7]$? In the solution I'm reading they say that $t=\frac{t^8}{t^7}$ but I don't see why that's relevant. Does what I've said work?
Show that $\mathbb Z[t^4,t^7]$ is not integrally closed in $\mathbb Z[t]$
commutative-algebraintegral-dependence
Best Answer
Recall an element of $S \supset R$ is called integral (over $R$) whenever it satisfies some monic polynomial in $R[x]$.
Then $R$ is called integrally closed in $S$ if it already contains all the elements which are integral over it. If we don't provide $S$ when we talk about integral closure, then $S$ is assumed to be the field of fractions of $R$.
You're entirely right that
so $t$ witnesses the fact that $\mathbb{Z}[t^4, t^7]$ is not integrally closed in $\mathbb{Z}[t]$.
As for the $t = \frac{t^8}{t^7}$ comment, I assume that's to show that $t$ still shows the failure of integral closure when we talk about the field of fractions. It's not immediately clear that $t \in \mathbb{Q}(t^4, t^7)$, which is the field of fractions of our ring. But writing $t = \frac{t^8}{t^7} = \frac{(t^4)^2}{t^7}$ makes this obvious.
I hope this helps ^_^