Suppose $F^{\prime}/F$ is an algebraic extension of fields, and that $F^{\prime}$ is a finite field extension of $K^{\prime}(x^{\prime})$, where $x^{\prime}$ is transcendental over $K^{\prime}$, and that $F$ is a finite extension of $K(x)$, where $x$ is transcendental over $K$, and finally that $K\subseteq K^{\prime}$. We also know that the set of elements of $F^{\prime}$ that are algebraic over $K^{\prime}$ is equal to $K^{\prime}$, and that the set of elements of $F$ that are algebraic over $K$ is equal to $K$. So we have the following diagram:

$$\begin{matrix} & & F^{\prime} \\ & \diagup & | \\ F & & K^{\prime} \\ | & \diagup & \\ K & & \end{matrix}$$

And we know that $F^{\prime}/F$ is algebraic. What I want to show is that $K^{\prime}/K$ is also algebraic. I spent a couple of hours trying to prove it, but I didn't succeed.


Let us call $L:=K(x)$ and $L':=K'(x')$. First I show that $L'/L$ is an algebraic extension. It is easy, take $u\in L'$ then $u\in F'$ and because $F'/F$ is algebraic $F(u)=F[u]$ is of finite dimension over $F$. Hence it is of finite dimension over $L$ because :

$$[F[u]:L]=[F[u]:F][F:L]<+\infty $$

The first dimension is finite because of what I have wrote before and the second dimension is finite by hypothesis.

Finally $L[u]\subseteq F[u]$ hence $[L[u]:L]$ is finite dimensional.

Let us take now $u\in K'\subseteq L'$ because $L'/L$ is algebraic, we have a polynomial $P\in L[Y]$ such that $P\neq 0$ and $P(u)=0$. Now we want to write carefully the polynomial $P$ :

$$P=\sum_{n=0}^da_n(x)Y^n $$

where $a_n\in L=K(x)$ that is :

$$a_n(x)=\frac{Q_n(x)}{D_n(x)} $$

Where $Q_n$ and $D_n$ are two polynomials in $x$ with coefficients in $K$. Write $D:=lcm((D_n(x))_n)$ then :

$$D(x)P(Y)=\sum_{n=0}^{d}b_n(x)Y^n $$

where $b_n\in K[x]$ for all $n$. Write (for $t=max(deg(b_n))$ :

$$b_n=\sum_{l=0}^tk_{n,l}x^l $$

Then we have :

$$D(x)P(Y)=\sum_{n=0}^{d}\sum_{l=0}^tk_{n,l}x^lY^n $$

$$D(x)P(Y)=\sum_{l=0}^t(\sum_{n=0}^{d}k_{n,l}Y^n)x^l $$

Let us write this :

$$D(x)P(Y)=\sum_{l=0}^tS_l(Y)x^l $$

where : $S_l(Y)\in K[Y]$. We know that $D(x)P(u)=0$ and we want to deduce from this that for all $l$, $S_l(u)=0$, hence it suffices to show that $(1,x,x^2,...)$ is free over $K'$. We will do this by contradiction.

Assume that $K'(x)$ is of finite dimension over $K'$. One has $L(x')$ is of finite dimension over $L$ (because $L'/L$ is algebraic) so one can tensor over $K$ by $K'$ to get that :

$$L'=L(x')\otimes_K K'\text{ is of finite dimension over } L\otimes_K K'=K'(x) $$

Now we have :

$$[L':K']=[L':K'(x)][K'(x):K'] $$

Hence, if we had that $[K'(x):K']$ is finite, one would have that $x'$ cannot be transcendental over $K'$ which is a contradiction. Hence $(1,x,x^2,...)$ is a family in $L'$ free over $K'$ which leads to : $S_l(u)=0$ for all $l$. If all polynomials $S_l\in K[Y]$ were null then $P$ itself would be null, because $P$ isn't, at least one of the $S_l$'s cannot be null and hence gives a non-trivial cancelling polynomial for $u\in K'$ over $K$. Hence $K'/K$ is algebraic as well.

  • $\begingroup$ There are only a few things I didn't understand. What is the $ppcm((D_n(x))_n)$? And what does it mean to be free over $K^{\prime}$ (it's the same of being free as a module, i.e., in that case it means that $\{1,x,x^2,\ldots\}$ is a basis for $K^{\prime}$ over $K$)? The rest of your demonstration is clear enough! $\endgroup$ – Larara May 11 '15 at 15:35
  • $\begingroup$ I guess that the ppcm reads "le plus petit commun multiple" (once, by your name, you're probably french)? $\endgroup$ – Larara May 12 '15 at 2:14
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    $\begingroup$ Yes, I mean $(1,x,x^2,...)$ is free as a familiy in a $K'$-vector space. It almost clear that this a basis for $K'[x]$ over $K'$. Your guess about $ppcm$ is right it does mean $lcm$, i.e. "least common multiple" and yes it is a Gallicism (I will edit it). $\endgroup$ – Clément Guérin May 12 '15 at 6:48

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