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If $v$ is algebraic over $K(u)$ for some $u\in F$, $F$ is an extension over $K$, and $v $ is transcendental over $K$, then $u$ is algebraic over $K(v)$.

I came across this problem in the book Algebra (Hungerford), I feel it shouldn’t be too hard but I just can’t work it out. Hope you can help me. Thank you!

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You can assume that $F=K(u,v)$. Let $X$ and $Y$ be independent indeterminates over $K$.

Since $v$ is algebraic over $K(u)$, there exist $a_i,b_i\in K[X]$ with

$$

\frac{a_0(u)}{b_0(u)}+

\frac{a_1(u)}{b_1(u)}v+

\dots

\frac{a_{n-1}(u)}{b_{n-1}(u)}v^{n-1}+

v^n=0

$$

with $n\ge0$. If $n=0$, then $u$ is algebraic over $K$, hence also over $K(v)$. Assume $n>0$ and clear the denominators to get

$$

f_0(u)+f_1(u)v+\dots+f_n(u)v^n=0

$$

with $f_i\in K[X]$.

Since $v$ is transcendental over $K$, you know that at least one $f_i(u)\in K(u)\setminus K$, in particular one of the polynomials $f_i(X)$ has positive degree.

Now consider the polynomial $g(X,Y)=f_0(X)+f_1(X)Y+\dots+f_n(X)Y^n$, that we can rewrite as $g(X,Y)=g_0(Y)+g_1(Y)X+\dots+g_m(Y)X^m$, with $g_i(Y)\in K(Y)$, and where $m>0$ because of the remark above.

Since $0=g(u,v)=g_0(v)+g_1(v)u+\dots+g_m(v)u^m$, we have proved the claim.

Hint: $v$ algebraic over $K(u)$ $\implies$ $v$ is root of a polynomial with coefficients in $K(u)$ $\implies$ $v$ is root of a polynomial with coefficients in $K[u]$ $\implies$ for some two variable polynomial $P\ne 0$ with coefficients in $K$, $P(u,v) = 0$. Expanding this in powers of $u$:

$$0 = P(u,v) = P_0(v) + P_1(v)u + \cdots + P_n(v)u^n,$$

i.e., $u$ is root of a polynomial with coefficients in $K[v]$. If $v$ were algebraic over $K$, all the $P_i(v)$ could be zero while $P\ne 0$. As $v$ is transcendental. this is impossible.

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