Prove that $E[F]$ is not a field. This is a question from Lang's algebra:
Let $E, F$ be extensions of a field $k$, contained in some bigger field $L$. We
can form the ring $E[F]$ generated by the elements of $F$ over $E$.
It is clear that the elements of $E[F]$ can be written in the form $a_1 b_1 + \dots + a_n b_n$ with $a_i \in E$ and $b_i \in F$.

*

*How are we claiming that it is not a field? I can prove all the properties of a ring. But I can't prove/disprove that it is a not a field.

*What will be the quotient field of this ring?

 A: It certainly can be a field, for example if

*

*$E\subseteq F$ or $F\subseteq E$ or

*The elements of $F$ are algebraic over $E$.

An example when $E[F]$ is not a field: Take $E=\Bbb Q(x),F=\Bbb Q(y)$ inside $L=\Bbb Q(x,y)$. Then $E[F]$ consists of those rational functions whose denominator can be written in the form $p(x)q(y)$ with $p(x)\in \Bbb Q[x], q(y)\in \Bbb Q[y]$. Hence $x+y$ doesn't have an inverse in $E[F]$, so $E[F]$ is not a field.
The field of fractions of this ring will be the compositum $EF$ of $E$ and $F$ in $L$. It consists of the elements of the form $$\frac{a_1b_1+\dots+a_nb_n}{c_1d_1+\dots+c_md_m}$$
where $a_i,c_j\in E, b_i,d_j\in F$.
Edit: Assume that the elements of $F$ are algebraic over $E$, then all elements in $E[F]$ are also algebraic over $E$, so if $0\ne x\in E[F]$, there are some $a_1,\dots,a_n$ with $$x^n+a_1x^{n-1}+\dots+a_n=0$$
We may assume $a_n\ne0$. We get: $$x^{-1}=(-a_n)^{-1}(x^{n-1}+a_1x^{n-2}+\dots+a_{n-1})\in E[F]$$ Hence $E[F]$ is a field (alternatively this follows from the fact that if $\alpha$ is algebraic over $E$, then $E[\alpha]=E(\alpha)$ is already a field).
As the elements of $E[F]$ are of the form $a_1b_1+\dots+a_nb_n$ it is clear that the field of fractions $Q$ contains the elements of the form $$\frac{a_1b_1+\dots+a_nb_n}{c_1d_1+\dots+c_md_m}$$ As the set of elements of this form is already closed under taking inverses (in $L$) it follows that this is already a field, hence it is the field  of fractions of $E[F]$. As this is a field that contains both $E$ and $F$ it contains the compositum $EF$. On the other hand, since $EF$ contains $E,F$ and is a field it also has to contain those elements of the form $\frac{a_1b_1+\dots+a_nb_n}{c_1d_1+\dots+c_md_m}$, therefore $Q$ is a subfield of $EF$. It follows $EF=Q$.
