I am looking for the minimum degree polynomial with integer coefficients that has $\sqrt[n]{a}$ as a root where $n$ and $a$ are integers, and where $a>1$ is not a perfect power. I believe that the minimum degree is $n$, for example $x^{n} - a$ works, but I can't find a way to prove that the degree cannot be less than $n$. Based on my research, it seems like you can prove it through some theorems in Field Theory(like this Minimal degree of polynomial with $a^{1/b}$ as root) but I am looking for a proof that doesn't require that, and uses just basic algebra.
What is the minimum degree of polynomial with $\sqrt[n]{a}$ as a root that has integer coefficients?
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5The minimal polynomial of $\sqrt[6]{8}$ is not $x^6-8$. You need to add more conditions on $a$ and $n$. – jjagmath Feb 23 '24 at 14:31
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Would it be better to say n cannot be a multiple of 4 and $\sqrt[n]{a}$ cannot be simplified? – Bob TheBuilder Feb 23 '24 at 15:11
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Been simplified has not a precise mathematical meaning. If you "simplify" $\sqrt[6]{72} $ to $\sqrt{2}\sqrt[3]{3}$ then what is $n$ now? – jjagmath Feb 25 '24 at 12:07
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That is a good point. Perhaps a definition could be that $\sqrt[n]{a}$ can be simplified if there exists $m$ and $b$ such that $\sqrt[n]{a} = \sqrt[m]{b}$. By this definition, $\sqrt[6]{72}$ cannot be simplified, but $\sqrt[6]{8}$ can be because $\sqrt[6]{8} = \sqrt{2}$. – Bob TheBuilder Feb 25 '24 at 17:02
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So $\sqrt 2$ can be simplified? Because there exists $m=6$ and $b=8$ such that $\sqrt 2 = \sqrt[m] b$ – jjagmath Feb 25 '24 at 18:10
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That is true, if we change the definition to require $m < n$, can you see any other issues? Thanks for pointing these mistakes out. – Bob TheBuilder Feb 26 '24 at 03:50
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You can simply write $\sqrt[n] a$ where $a>1$ is not a perfect power. – jjagmath Feb 26 '24 at 14:52
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I just edited the question to add that in. Thanks again. – Bob TheBuilder Feb 26 '24 at 14:57