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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?

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3  
Do you know what "integral closure" means? –  Chris Eagle Mar 16 '13 at 23:27
    
I thought it meant the following: The integral closure of $A$ is every element in the fraction field of $A$ which is integral over $A$. –  Thomas Mar 16 '13 at 23:32
    
So why do you believe that your $y$ must be in the integral closure of $A$? –  Chris Eagle Mar 16 '13 at 23:35
    
Because $y$ is in the fraction field $F(x)$ and I have found a polynomial $p(t)=(x^2-5)t-(x^3+2)$ with coefficients in $A$ such that $p(y)=0$, i.e. $y$ is integral over $A$. –  Thomas Mar 16 '13 at 23:37
1  
So you don't know what "integral over $A$" means. Better look it up. –  Chris Eagle Mar 16 '13 at 23:38

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