Integral closure of rational polynomial ring

[pantboio]

Consider the ring $\mathbb{Q}[X]$ of polynomials in $X$ with coefficients in the field of rational numbers. Consider the quotient field $\mathbb{Q}(X)$ and let $K$ be the finite extension of $\mathbb{Q}(X)$ given by $K:=\mathbb{Q}(X)[Y]$, where $Y^2-X=0$.Let $O_{K}$ be the integral closure of $Q[X]$ in $K$. Certainly $O_{K}$ contains both $\mathbb{Q}[X]$ and $Y$, hence
$$ O_{K}\supseteq \mathbb{Q}[X][Y]$$
My guess is that actually "=" holds. How can be proved this?
Thank u all in advance

 

[Aaron Bex]

You can prove it by showing that $\mathbb{Q}[X][Y]$ is integrally closed.
It suffices to prove that $\mathbb{Q}[X][Y]$ is integrally closed in its quotient field, which is $K$.
Let $\alpha \in K$ be integral over $\mathbb{Q}[X][Y]$. Since $K$ is a finite extension over $\mathbb{Q}(X)$, we can write $\alpha$ as
$$\alpha = b_0 + b_1 Y + \cdots + b_n Y^n$$
where $b_i \in \mathbb{Q}[X]$ for all $i$.
By writing each $b_i$ as a polynomial in $X$, and then multiplying the whole equation by $X^N$ for some sufficiently large $N$, we can assume that all the coefficients of the polynomials are integers.
Now, let $f$ be the minimal polynomial of $\alpha$ over $\mathbb{Q}[X]$. Since all the coefficients of the above expression are integers, it follows that $f$ has integer coefficients.
But since $Y^2 - X = 0$, this implies that $f$ must have degree at most $2$. Moreover, since $\alpha \