# 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 $$b_i \in F$$.

1. 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.
2. What will be the quotient field of this ring?

• $$E\subseteq F$$ 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 $$EF=Q$$.
• (Q-1) Can you prove that if the elements of $F$ are algebraic over $E$, then $E[F]$ is a field? (Q-2) I was not able to prove why elements of the form $\frac{a_1b_1+\dots+a_nb_n}{c_1d_1+\dots+c_md_m}$ will be the compositum $EF$ (i.e. the smallest field containing $E$ and $F$)? How do we prove that the collection of elements is $EF$ and not some other field that contain $E$ and $F$? Commented Jan 30, 2021 at 8:57