show if $X:Z$ and $Y:X$ are algebraic extensions then $Y:Z$ is algebraic

Question

I have found a proof but my proof assumes that the extensions are finite. I am trying to prove it whether or not the extensions are finite. Any help please.

Answer 1

Suppose $F/E/K$ is a tower of extensions with $F/E$ and $E/K$ algebraic, and let $u\in F$. We will show that $u$ is algebraic over $K$. To this end, note that $u$ is algebraic over $K$, so $u$ is a root of some polynomial $$f(x)=a_0+a_1x+\ldots+a_nx^n\in E[x].$$ In particular, $a_0,\ldots,a_n$ are algebraic over $K$, so $u$ is algebraic over the finite extension $K(a_0,\ldots,a_n)/K$. Now, $K(a_0,\ldots,a_n,u)/K$ is finite (since it is finite over $K(a_0,\ldots,a_n)$), hence $u$ is algebraic over $K$.

[This proof above is pictured on the cover of the Beachy-Blair Abstract Algebra textbook.]

Answer 2

You can prove that $[Y:Z]\le [Y:X][X:Z]$ by noting $y\in Y$ can be represented as

$$y=\sum_{i=1}^{[Y:X]}\left(\sum_{j=1}^{[X:Z]} c_{ij}z_{ij}j\right)$$

So the size of a spanning set does not exceed the finite number $[Y:X][X:Z]$. Here we choose $c_{ij}$ and $z_{ij}$ such that

$$\sum_{j=1}^{[X:Z]}c_{ij}z_{ij}=c'_ix_i$$

and

$$\sum_{i=1}^{[Y:X]}c'_ix_i=y$$

To see each of the others is finite, it's a simple matter of noting that if $y\in Y$ then it is in some finite extension of $X$, eg $X(y)$, which we know is finite because $y$ satisfies a minimal polynomial over $X$. But then if we have

$$y=\sum_{i=1}^{[X(y):X]}c_i'x_i$$

we may choose $z_{ij}\in Z$ satisfying the condition from above because each $x_i$ is algebraic, so $Z(x_1,x_2,\ldots, [X(y):X]$ generate a finite extension of $Z$.

So you reduce to the case $X=Z(x_1,\ldots, x_{[X(y):X]}$ and $Y=X(y)$ which are finite extensions and the general case follows.

Answer 3

Let $F\subset K$ and $K\subset L$ be algebraic extensions of fields, and pick $\alpha\in L$.

$\alpha$ is algebraic over $K$ by assumption, so it satisfies a polynomial equation $\sum_{i=0}^n c_i \alpha^i = 0$ with $c_i\in K$ and $c_n\neq 0$. In particular, $\alpha$ is algebraic over $F(c_1, \ldots , c_n)$.

Now consider the extension of fields $F \subset F(c_1, \ldots , c_n) \subset F(\alpha, c_1, \ldots , c_n)$. Both extensions are finite, so we can apply the result for finite extensions to conclude that $\alpha$ is algebraic over $F$.