Quadratic extension of $\mathbb{Q}(X)$

Question

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 a quadratic extension of $\mathbb{Q}(X)$ given by $K:=\mathbb{Q}(X)[d]$.

Let $O_{K}$ be the integral closure of $\mathbb Q[X]$ in $K$. Certainly $O_{K}$ contains both $\mathbb{Q}[X]$ and $d$, hence $$ O_{K}\supseteq \mathbb{Q}[X][d].$$ Do you think the previous inclusion is strict, or the equality holds?

EDIT. The conclusion $O_{K}\supseteq \mathbb{Q}[X][d]$ is false. So the question is: who is $O_{K}$?

@Qiaochu Yuan the root of a monic irreducible polynomial of degree 2 with coefficients in $\mathbb{Q}(X)$ — Federica Maggioni, Mar 13 '13 at 19:02
For example, we could take $,d,$ to be such that $,d^2=X\iff d,$ is a root of $,p(t):=t^2-X\in\Bbb Q[x][t],$...? — DonAntonio, Mar 13 '13 at 19:09
@DonAntonio i've already considered that case in another question http://math.stackexchange.com/questions/329284/integral-closure-of-mathbbqx-in-mathbbqxy?noredirect=1#comment711757_329284 — Federica Maggioni, Mar 13 '13 at 19:18
@DonAntonio now i'm considering a generic quadratic extension of $\mathbb{Q}(X)$ — Federica Maggioni, Mar 13 '13 at 19:19

score 5 · Accepted Answer · edited Apr 13 '17 at 12:20

You write: "Certainly $O_{K}$ contains both $\mathbb{Q}[X]$ and $d$, hence $ O_{K}\supseteq \mathbb{Q}[X][d]$"

But this is not true:
Take $d=\sqrt \frac {1}{X}$, a root ot $T^2-\frac {1}{X}\in \mathbb Q(X)[T]$.
Then $d$ is not integral over $\mathbb Q[X]$ because if it were, then also $d^2=\frac {1}{X}$, would be integral over $\mathbb Q[X]$ and we both know very well that this is absurd.
So, $d\notin O_K$ and thus $ O_{K}\supseteq \mathbb{Q}[X][d]$ does not hold.

Edit
As for the computation of $O_K$, here is a result which may be of help: it is given as Theorem 9.2 (page 65) in Matsumura's Commutative ring theory.

Theorem
Let $A$ be an integrally closed domain and $K$ an algebraic extension of its field of fraction $F=Frac(A)$.
Then an element $k\in K$ is integral over $A$ iff its minimal polynomial over $F$, the monic polynomial $f(T)=Irr(k,F,T)\in F[T]$, has its coefficients in $A$.

In your case, if you suppose for example that $d=\sqrt {P(X)}$ with $P(X)\in \mathbb Q|X]$ a square-free polynomial, the theorem above will allow you to prove that $O_K$ is the ring $\mathbb Q[X][d]$ (details in Matsumura).
(The above class of examples generalizes the result of your preceding question but doesn't completely solve the problem of calculating $O_K$ for general quadratic extensions $K$ of $\mathbb Q(X)$).

whoops, you are completely right, so how can i compute $O_{K}$ in this case? — Federica Maggioni, Mar 13 '13 at 19:49
Dear Federica, first let me congratulate you on your comment: everybody makes mistakes but it takes laudable maturity to quickly recognize one's error. I'll write a little edit addressing the computation of $O_K$. — Georges Elencwajg, Mar 13 '13 at 20:05
I think the most important question here is not the computation of $O_K$ which can turns out to be cumbersome. It would be very nice to know when $O_K=R[d]$, where $R=\mathbb Q[X]$ (or, in general, an integral domain.) The answer is known for $R$ a Dedekind domain and says the following: $O_K=R[d]$ iff the minimal polynomial of $d$ over $K$ (the field of fractions of $R$) doesn't belong to $M^2$ for all maximal ideals $M$ of $R[X]$. — , Mar 13 '13 at 23:27

Quadratic extension of $\mathbb{Q}(X)$

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