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How can I show that if $a,b$ are elements of a field $K$ and $k, l$ are element of the positive natural numbers, then the element $\sqrt[k]{a}\cdot\sqrt[l]{b}$ is also algebraic over $K$.

I'm stuck in the step: $x^{kl} - a^l \cdot b^k = 0$. So how can I now show that such a polynomial exists in $K$?

And further I have a problem with the following: if $a$ is algebraic over $K$, then for all $n$ (out of the natural numbers) the element $\sqrt[n]{a}$ is algebraic over $K$ as well. How can I show that?

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Both parts are straightforward.

$x^{kl} - a^l \cdot b^k$ is clearly a polynomial with coefficients in $K$, as $a, b \in K$.

If $a$ is a root of $f(x) \in K[x]$, then $\sqrt[n]{a}$ is a root of $g(x) = f(x^{n})$.

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  • $\begingroup$ So a more extended way to say the 2nd part would be: f(x) = 0 and x = a, so f(a) = 0 and therefore a is a root of f(x). Further f(x^n) = 0 and x is n-root(a) --> f(n-root(a) ^n) = f(a) = 0 and therefore its also a root. right? $\endgroup$ – user122299 Jan 18 '14 at 23:09
  • $\begingroup$ @user122299, please be careful. The polynomial $f(x) \in L[x]$ is by definition non-zero. It is only its evaluation at $a$ which is zero, that is, $f(a) = 0$. Then $f\left((\sqrt[n]{a})^{n}\right) = f(a) = 0$. And, please, try and use $\LaTeX$: see the editing that has been done to your post the see the proper way of writing roots etc. $\endgroup$ – Andreas Caranti Jan 19 '14 at 8:48

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