Vtyjkyuk

While playing around with fraction fields I wanted to proof a statement in a special case. But somehow I ended up with a proof that doesn't use the properties of the special case. Now I know that the general case does not hold true, hence there must be an error in my proof.

Let me begin with some background and intuition on the topic, based on the following definitions:

Let $R$ be a domain (i.e. a commutative ring with identity having no nonzero zero-divisors). Then

(i) $Rleft[etaright]=a+beta : a,b in R$

(ii) $Rleft(etaright)=fracfg : f,g, in Rleft[eta right]$

where $eta in S$ fore some domain $S supseteq R$.

(EDIT: as rschwieb pointed out in the comments, this is not the standard definition. Rather $Rleft[etaright]=spanleft(1_R,etaright)$. These definitions can also be stated as follows:

(i) $Rleft[etaright]$ is the smallest ring containing $R$

(ii) $Rleft(etaright)$ is the smallest field containing $R$)

There are some situations where these two definitions coincide, e.g. for $R=mathbbQ$ and $eta=sqrtz$ for some $z in mathbbZ$. Then one has
$$mathbbQleft(sqrtzright) simeq mathbbQleft[sqrtzright]$$

(in this case even equality holds)

proof

A priori one finds the inclusion $mathbbQleft[sqrtzright] subseteq mathbbQleft(sqrtzright)$. For the other direction, let $k in mathbbQleft(sqrtzright)$. Then $k$ is of the form $$k=fracfg=fraca+bsqrtzc+dsqrtz$$ for $a,b,c,d in mathbbQ$. But we can find $r,s in mathbbQ$ s.t. $$fraca+bsqrtzc+dsqrtz=r+ssqrtz in mathbbQleft[sqrtzright]$$ namely by the conditions $rc+zsd=a$ and $rd+sc=b$, which we can decouple and solve for $s$ and $r$. Hence $mathbbQleft(sqrtzright) = mathbbQleft[sqrtzright]$.

(EDIT
In this proof I used the definition $Rleft[etaright]=a+beta : a,b in R$, not the standard one! When using the standard definition, one can adapt the proof by letting $k=fracsum_i=0^n a_i+b_i z^i/2sum_j=0^m c_i+d_i z^i/2$ and one would eventually find that - as rschwieb mentioned - we need $sqrtz$ to be algebraic.)

It seems to me that $Kleft(etaright) simeq Kleft[etaright]$ should hold for any field extension $L/K$ as long as $exists kappa in L$ s.t. $kappa^2=eta in L$ (the case $eta in K$ is trivial); we only need inverses to guarantee a solution to the coupled linear system of equations. Is this assumption correct and - if so - is there a way to classify the fields having this property?

As mentioned in the beginning, I tried to prove this more general case by constructing an isomorphism, but I nowhere used the property of $eta$ being a square, so I think my proof must be wrong. Any hint on where I made a mistake is very much appreciated.

proof

Let $Phi: Kleft(etaright) to Kleft[etaright]$ by $fracfg mapsto f$. This is a homomorphism of rings, since $$Phileft(fracfg+frachjright)=Phileft(fracfj+ghgjright)=fj+gh=Phileft(fracfj1right)+Phileft(fracgh1right)=Phileft(fracfjjgright)+Phileft(fracghgjright)=Phileft(fracfgright)+Phileft(frachjright)$$ and $$Phileft(fracfgcdotfrachjright)=fh=Phileft(fracfgright)cdotPhileft(frachjright)$$
Trivially $Phi$ is surjective, and if $fracfg in kerPhi$, then $f=0$ and so is $fracfg$. Hence $Phi$ is an isomorphism.

(Caveat: this is written with the standard meaning of $k[eta]$ and $k(eta)$ in mind, which may not have been the poster's intent.)

It seems to me that $Kleft(etaright) simeq Kleft[etaright]$ should hold for any field extension $L/K$ as long as $exists kappa in L$ s.t. $kappa^2=eta in L$ (the case $eta in K$ is trivial); we only need inverses to guarantee a solution to the coupled linear system of equations. Is this assumption correct and - if so - is there a way to classify the fields having this property?

In general $K(eta)=K[eta]$ iff $eta$ is an element of $L$ which is algebraic over $K$. The degree of the minimal polynomial of $eta$ over $K$ is the degree of the field extension.

If you are particularly interested in cases where $|K(eta):K|=2$ then you are looking at quadratic extensions of $K$. This will occur exactly when $eta^2$ is in the span of $1,eta$.

The condition you gave (that $eta$ is a square in $L$) does not suffice: for example, $pi$ has a square root in $mathbb C$, but $mathbb Q[pi]neq mathbb Q(pi)$

In particular, could you modify this proof by using the fact from your answer, that η needs to be algebraic over $K$?

The proof you have undertaken, which aims to show $k[eta]$ and $k(eta)$ are isomorphic is making things too hard. The two sets are actually equal. You already noted that $k[eta]subseteq k(eta)$, so we just need to prove the other containment. It suffices to show $k[eta]$ is a field.

If $eta$ is algebraic over $K$, then there is necessarily an irreducible polynomial $p(x)in k[x]$ for which $eta$ is a root. The evaluation map $k[x]to k[eta]$ has precisely $I=(p(x))$ as its kernel. (This would be a good exercise for you.)

Therefore $k[x]/(p(x))cong k[eta]$ by the first homomorphism theorem. But since $p(x)$ is irreducible, it generates a maximal ideal of $k[x]$, so the left hand side is a field (and so the right hand side is a field also.)

But you already defined $k(eta)$ as the smallest field containing $k$ and $eta$, so $k(eta)subseteq k[eta]$.