Algebraic Elements and Transcendental Elements
Proposition 1. Let $E=F(u)$, where $u$ is transcendental over $F$. If $K\neq F$ and $K$ is an intermediate field of $E/F$, then $u$ is algebraic over $K$.
Proof: Take any
$$
v\in K\setminus F.
$$
Since
$$
v\in F(u),
$$
there exist $f(x),g(x)\in F[x]$ such that
$$
v=\frac{f(u)}{g(u)}.
$$
Therefore,
$$
h(x)=vg(x)-f(x)
$$
is a polynomial over $F(v)$ satisfying
$$
h(u)=0.
$$
Moreover, since
$$
v\notin F,
$$
the leading coefficient of $h(x)$ is nonzero, so
$$
h(x)\neq 0.
$$
Hence $u$ is algebraic over $F(v)$, and therefore also algebraic over $K$.
Proposition 2. Let $E$ be an extension field of $F$, let $a\in E$, and let $b\in F(a)$ with
$$
b\notin F.
$$
Then:
If $a$ is algebraic over $F$, then $b$ is also algebraic over $F$.
If $a$ is transcendental over $F$, then $b$ is also transcendental over $F$.
Proof:
For part (1), the conclusion is immediate.
We now prove part (2). By Proposition 1, $a$ is algebraic over $F(b)$, where we take
$$
K=F(b),\qquad E=F(a).
$$
Hence
$$
[F(a):F(b)]<\infty.
$$
On the other hand, since $a$ is transcendental over $F$, we have
$$
[F(a):F]=\infty.
$$
Therefore, from the dimension formula
$$
[F(a):F]=[F(a):F(b)][F(b):F],
$$
it follows that
$$
[F(b):F]=\infty.
$$
Hence $b$ is also transcendental over $F$.
Proposition 3. Let $E$ be an extension field of $F$, and let $a,b\in E$. If $a$ is transcendental over $F$, but algebraic over $F(b)$, then $b$ is algebraic over $F(a)$.
Proof: Let $p(x)$ be the minimal polynomial of $a$ over $F(b)$, and suppose that its degree is $n$. Then
$$
p(a)=0,
$$
and there exist $f_i(x),g_i(x)\in F[x]$ such that
$$
\frac{f_1(b)}{g_1(b)}a^n+\cdots+\frac{f_n(b)}{g_n(b)}a+\frac{f_{n+1}(b)}{g_{n+1}(b)}=0.
$$
After clearing denominators and taking the numerator, we obtain a nonzero polynomial
$$
u(x,y)\in F[x,y]
$$
such that
$$
u(a,b)=0.
$$
Now define
$$
v(y)=u(a,y).
$$
Then $v(y)$ is a polynomial over $F[a]$ satisfying
$$
v(b)=0,
$$$v(y)$ is not the zero polynomial.
The second assertion holds because $a$ is transcendental over $F$.
Therefore $b$ is algebraic over $F(a)$.
The cover image of this article was taken in Singapore.
Algebraic Elements and Transcendental Elements