Question: Let $F$ be a field and $A\subset F[x]$ the polynomials without the linear term. Prove that $F[x]$ is the integral closure of $A$.
My Proof:
Since we have $x=x^3/x^2$, the field of fractions of $A$ is $F(x)$, because $x^2,x^3\in A$. Also, $x\in F(x)$ is a root of $t^2-x^2\in A[t]$, so $A$ is not integrally closed. In fact, $F[x]$ is generated by $1,\ x$ as an $A$-module, so any element of $F[x]$ is integral over $A$. And since every UFD is integrally closed, $F[x]$ is closed, hence it must be the integral closure of $A$.
But, when I take $y=\frac{x^3+2}{x^2-5}\in F(x)$, then $(x^2-5)\cdot t- (x^3+2)\in A[t]$ has $y$ as a root. So $y$ should be in the integral closure of $A$. What is my mistake?
Best Answer
From the comments above.
If $A \subset B$ are commutative rings with identity, then an element $\alpha \in B$ is said to be integral over $A$ if $\alpha$ satisfies a monic polynomial $f(X) \in A[X]$, that is, a polynomial whose leading coefficient is $1$, so $f$ is of the form $$ f(X) = X^n + a_{n-1}X^{n-1} + \dots + a_1 X + a_0 $$ for some $a_i \in A$, $0 \leq i \leq n-1$.
So, since $p(t) = (x^2 - 5) \cdot t - (x^3 + 2) \in A[t]$ is not monic, $y = \frac{x^3 + 2}{x^2 - 5}$ is not integral over $A$.