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Let there be 3 fields $A$, $B$ and $C$.

If all elements of $A$ are algebraic over $B$ and all elements of $B$ are algebraic over $C$, prove that this implies that all elements of $A$ is algebraic over $C$.

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    $\begingroup$ If you plug a polynomial in $t$ in for every value of $x$ in polynomial $f(x)$, the result is a polynomial in t. $\endgroup$ – vadim123 Oct 9 '14 at 16:55
  • $\begingroup$ @vadim123 I'm not sure if the composition of polynomial $f$ with coefficients in $B$ and polynomial $g$ with coefficients in $C$ which is $g(f(x))$ have coefficients in $C$ $\endgroup$ – MathsMy Oct 9 '14 at 17:15
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Let it be that element $\alpha\in A$ has minimal polynomial $\sum_{i=0}^{n}\beta_{i}X^{i}$ with $\beta_{i}\in B$.

Then $C\subset B_{0}:=C\left(\beta_{0},\dots,\beta_{n}\right)$ and $B_{0}\subset B_{0}\left(\alpha\right)$ are finite extensions.

Consequently $C\subset B_{0}\left(\alpha\right)$ is a finite (hence algebraic) extension.

This tells us that $\alpha$ is algebraic over $C$.


edit:

Used are the following lemma's:

  • If $K\subset L$ and $L\subset M$ are finite extensions then $K\subset M$ is a finite extension.

  • Every finite extension is algebraic. (A proof of that can be found here).

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If you have an element $\alpha \in A$ which is algebraic over $B$, what can you say about the form of $\alpha$? Similarly, if you have an element $\beta \in B$ which is algebraic over $C$, what can you say about the form of $\beta$? Can you somehow write $\alpha \in A$ as an algebraic element over $C$?

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  • $\begingroup$ If $\alpha \in A$ is algebraic over $B$, then there exists a polynomial with coefficients in $B$ such that it vanishes at $\alpha$. Similarly for $\beta \in B$ over $C$. I was thinking of composition of polynomials but I'm not sure if that gives a polynomials with coefficients in $C$. $\endgroup$ – MathsMy Oct 9 '14 at 17:10
  • $\begingroup$ @Dosomemaths Try it. You can say that $\alpha$ is algebraic over $B$, and every coefficient in the polynomial would be algebraic over $C$, and this composition of polynomials would give a new polynomial over $C$ with $\alpha$ as a solution. $\endgroup$ – Johanna Oct 9 '14 at 21:12

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