Let $E$ be an extension of field $F$, and let $\alpha, \beta \in E$. Suppose $\alpha$ is transcendental over $F$ but algebraic over $F(\beta)$.

Show that $\beta$ is algebraic over $F(\alpha)$.

Okay, first questions: What does the notation $F(\alpha)$ and $F(\beta)$ mean? And being transcendental means it solves no equations with rational coefficients, but what does it mean for a field?

  • $\begingroup$ I think transcendental has nothing to do with it. Just try this: assume /beta is degree 2 and /alpha also degree 2 and try that and you'll see how it works. $\endgroup$ – tomrlopes Mar 28 '14 at 22:01
  • $\begingroup$ @tomrlopes I don't know what to "try" though, cause I don't know what type of elements are in $F(\beta)$ besides $\beta$ and elements of $F$. $\endgroup$ – terrible at math Mar 28 '14 at 22:06
  • $\begingroup$ Everything in $F(/beta)$ is a polynomial in $/beta$ I can't say too much more or I'll give it all away. $\endgroup$ – tomrlopes Mar 28 '14 at 22:09
  • $\begingroup$ @tomrlopes The claim is false if $\alpha$ is algebraic over $F$. If $\alpha$ is algebraic over $F$, it also is algebraic over $F(\beta)$ whatever $\beta$ is. So just take any $\beta$ transcendental over $F(\alpha)$ (equivalently, in this case, over $F$) for a counterexample. E.g. $\beta = T$, a new variable. $\endgroup$ – Magdiragdag Mar 29 '14 at 11:18
  • $\begingroup$ Ah, so the polynomial for $\alpha$ over $F(\beta)$ is actually over F $\endgroup$ – tomrlopes Apr 5 '14 at 2:06

$F(\alpha)$ means the smallest field containing both $F$ and $\alpha$.

$\gamma$ algebraic over $F$ means that there is a non-zero polynomial $p(X) \in F[X]$ (i.e., a polynomial with coefficients in $F$) with $p(\gamma) = 0$. (And transcendental means such a polynomial does not exist).

Now the problem itself. The situation is as follows.

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Since $\alpha$ is algebraic over $F(\beta)$, there is a non-zero polynomial $f(X) \in F(\beta)[X]$ with $f(\alpha) = 0$. The coefficients are elements of $F(\beta)$, but clearing denominators we may as well assume they are elements of $F[\beta]$.

So, $f(\alpha)$ is a polynomial expression in both $\alpha$ and $\beta$ and we can see it as a polynomial expression $g(\beta)$ in $\beta$ with coefficients in $F[\alpha]$, i.e., $g(Y) \in F[\alpha][Y]$. (To be precise, there is a polynomial $h(X,Y) \in F[X,Y]$ such that $f(X) = h(X,\beta)$ and $g(Y) = h(\alpha,Y)$.) Now $0 = f(\alpha) = g(\beta)$.

What is still left to show is that $g(Y)$ is not the zero polynomial, i.e., that not all its coefficients are $0$. But its coefficients are of the form $c(\alpha)$ with $c(X) \in F[X]$ and because $\alpha$ is transcendental over $F$, $c(\alpha)$ is $0$ only if $c(X) = 0$. So, if $g(Y)$ were the zero polynomial, so would $f(X)$ be.

Example. Take $\alpha = T^2$ and $\beta = T^3$ in the field ${\mathbb Q}(T)$ of rational functions over ${\mathbb Q}$. Then $\alpha$ is transcendental over ${\mathbb Q}$. Also, $\beta$ is algebraic over ${\mathbb Q}(\alpha)$ as it satisfies $\beta^2 - \alpha^3 = 0$ (i.e., $\beta$ is a root of the polynomial $Y^2 - \alpha^3$ over ${\mathbb Q}(\alpha)$). Exactly the same relation shows that $\alpha$ is algebraic over ${\mathbb Q}(\beta)$ (as $\alpha$ is a root of the polynomial $\beta^2 - X^3$ over ${\mathbb Q}(\beta)$).

  • 2
    $\begingroup$ thank you and thank the guy below you, I am sure I can work with this. If somehow I can't, I will be back :) $\endgroup$ – terrible at math Mar 28 '14 at 19:46
  • $\begingroup$ here's what I have so far: $\alpha$ is algebraic over $F(\beta)$ so $\exists P(x) \in F(\beta)[x] $ such that $P(\alpha) = 0$, but we also know that $\alpha$ is transcendental over $F$ so this must mean our $P$ lies in $F(\beta)[x] \setminus F[x]$. This makes sense to me, but I'm having trouble reasoning what the coeffecients look like. Ideally they are all $\beta$ but I doubt that is the case? $\endgroup$ – terrible at math Mar 28 '14 at 21:22
  • $\begingroup$ The coefficients of $P$ are, at first, rational functions in $\beta$, but clearing denominators you might as well assume they are polynomials in $\beta$. Now you have a polynomial expression in both $\alpha$ and $\beta$ that equals $0$. Consider this as a polynomial in $\beta$ with coefficients in $F(\alpha)$. $\endgroup$ – Magdiragdag Mar 29 '14 at 7:24

$F(\alpha)$ is the smallest field that contains both $F$ and $\alpha$ inside $E$.. Being transcendental means, it solves no equations with coefficients in the given field, i.e. $\beta \in E$ is transcendental over $F(\alpha)$, if no polynomial with coefficients in $F(\alpha)$ has $\beta$ as root.


This is an old question, but I would like to share an alternative solution I came up with when I did this exercise.

The conditions in the question is equivalent to the following diagram:


Now we claim $F(\beta)/F$ is transcendental. Suppose not, then $F(\beta)/F$ is algebraic. Since composition of algebraic extensions is algebraic, $F(\alpha, \beta)/F$ is algebraic, which contradicts $\alpha$ being transcendental over $F$.


Hence $F(\beta)/F(\alpha)$ is transcendental.


This means $\alpha, \beta$ are both transcendental over $F$. So $F(\alpha), F(\beta) \cong F(x)$ for some intermediate $x$ $\implies F(\alpha) \cong F(x) \cong F(\beta)$


Since $F(\alpha) \cong F(\beta)$, we can regard $F(\beta)$ as degree $1$ extension of $F(\alpha)$. $F(\beta)/F(\alpha)$ is algebraic since all finite extensions are algebraic.


Finally, we use again the fact that composition of algebraic extensions is algebraic to conclude $F(\alpha, \beta)/F(\alpha)$ is algebraic.



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