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Problem

Let $F \leq E$ be an extension of fields, let $a \in E$ and let $b \in F$.

Show that $a$ is algebraic over $F$ if and only if $a + b$ is algebraic over $F$.

Thoughts

Is the best approach here to try and explicitly construct some polynomial $f(x) \in F[x]$ that satisfies $f(a+b)=0$ (to show the $\Rightarrow$ direction), or is it better to try and argue for the finiteness of $F(a+b)$? [I understand these are effectively the same thing].

And for the reverse direction ($\Leftarrow$), is it best to argue that, since $b \in F$, $F(b)=F$ and so $F(a+b)=F(a)$ (I'm not sure if this can simply be stated without justification) meaning that $|F(a):F|=|F(a+b):F|=n$, where n is the degree of the minimmum polynomial for $a+b$, which exists by assumption.

Thinking about another problem

How can we change approach to deal with the equivalent problem but with $ab$ rather than $a+b$. Does the $\iff$ relation still hold? (We should probably add the assumption that $b \neq 0 $ for this case).

Any help is very appreciated. Thanks!

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If you have ideas, it is usually better to try those ideas to see where they lead than to become paralyzed with indecision about what to try. –  Hurkyl Jan 27 '13 at 23:49
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2 Answers

up vote 2 down vote accepted

$F(a+b)$ is the smallest subfield of $E$ that contains both $F$ and the element $a+b$. Since $b\in F$ it follows that $-b\in F$ and so $b=(a+b)-b\in F(a+b)$. It follows that $F(a+b)$ also contains $a$ and so: $F(a)\subseteq F(a+b)$, since $F(a)$ is the smallest subfield of $E$ containing both $F$ and $a$. Similarly, since $a\in F(a)$ and $b\in F$ it follows that $a+b\in F(a)$, and so $F(a+b)\subseteq F(a)$. So $F(a)=F(a+b)$ and the conclusion about algebraicity follows. A similar argument will work with multiplication instead of addition, as long as you are careful with the case $b=0$.

In general, find explicit minimal polynomials is hard. The trick of considering a minimal field instead usually simplifies a lot of proofs.

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I guess you're better off arguing the finiteness of $F(a+b)/F$: Clearly $a+b\in F(a)$ so that $F(a+b)\subset F(a)$ and finiteness is now clear. The same argument works for the reverse implication and the case of $ab$. In both cases the thing to note is that what we're proving is that $F(a+b)=F(a)$ and $F(ab)=F(a)$.

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