This question is from my assignment in commutative algebra.
Question: What is integral closure of $k[t^2 ,t^3]$ in $k[t]$, where $k$ is a field.
Attempt: k is a field and hence k is UFD , hence k[t] is UFD.
If C is the asked integral closure, then $C=${$x\in k[t]$: x is integral over $k[t^2, t^3]$}
So, $x^n +a_1 x^{n-1} +...+a_n=0$ , all $a_i$ $\in k[t^2,t^3]$. Now, $x^n +a_1 x^{n-1} +... + a_{n-1} x^1 = -a_n $. Now, x divide RHS implies x must divide LHS , so $x=k_0 + k_1 t +...+ k_n t^n $ divides $a_n$ but I am not able to move forward from this.
Please help.
As requested by OP:
Recall what you mean by the integral closure of $R$ in $S$.
This is the sub-ring containing every $s\in S$ such that $s$ is the root of a polynomial with all coefficients in $R$.
Now, let $R=k[t^2,t^3]$ and $S=k[t]$.
The polynomial $p(x)=x^2-t^2$ has $t$ as a root. Notice that $p(x)$ has all coefficients in $R$. See that $p(x)=1\cdot x^2+0\cdot x+(-t^2)\cdot1$?
Similarly, for any $j\in k$, the polynomial $x-j$ has $j$ as a root.
If you are given a sub-ring of $S$ containing both the whole $k$ and the element $t$, that sub-ring must be $S$ itself.