As you know, it holds that $\sqrt3\notin\mathbb{Q}(\sqrt2)$ The reason for this is about degree of field. That is, on the contrary, if $\sqrt3\in\mathbb{Q}(\sqrt2)$ then $\sqrt3+\sqrt2\in\mathbb{Q}(\sqrt2)$

So, $2$ = [$\mathbb{Q}(\sqrt2)$ : $\mathbb{Q}$] = [$\mathbb{Q}(\sqrt2)$ : $\mathbb{Q}(\sqrt2 +\sqrt3)$] [$\mathbb{Q}(\sqrt2 +\sqrt3)$ : $\mathbb{Q}$] This leads to the following. $deg(\sqrt2 +\sqrt3 , \mathbb{Q}$) = 2 However, this is a contradiction because we know that the degree of the irreducible polynomial which is $x^4-10x^2+1 \in \mathbb{Q}[x]$ is 4.

Now, lets consider following case,


If we can not present specific irreducible polynomials as in the previous situation(In fact, can not you calculate 9999 squared by hand?), what argument can we prove this fact??


In general, this type of problem can be quite difficult, though methods from Galois theory are often effective. However, in your particular case, there is a neat observation we can make. Indeed, the minimal polynomial of $2^{1/9999}$ over $\mathbb{Q}$ is $X^{9999}-2$, which is irreducible by Eisenstein at $2$. Hence, the degree of the extension $\mathbb{Q}(2^{1/9999})$ over $\mathbb{Q}$ is $9999$. If we had $\sqrt{3} \in \mathbb{Q}(2^{1/9999})$, then we would have $\mathbb{Q}(\sqrt{3}) \subset \mathbb{Q}(2^{1/9999})$, whence multiplicativity of degree gives us $[\mathbb{Q}(2^{1/9999}):\mathbb{Q}] = [\mathbb{Q}(2^{1/9999}):\mathbb{Q}(\sqrt{3})][\mathbb{Q}(\sqrt{3}):\mathbb{Q})]$. But $[\mathbb{Q}(\sqrt{3}):\mathbb{Q}] = 2$ does not divide $9999$, a contradiction. This type of reasoning is often quite useful, although again, these problems can be quite tricky in general: see this question, for example.

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  • $\begingroup$ Oh it is my mistake. So it would be better to think 10000 instead of 9999. $\endgroup$ – J.U.math Dec 6 '16 at 2:13
  • $\begingroup$ @Edgar.W: yes, since this sort of method would work for $1/n$ for any $n$ odd. I'm afraid I don't have any great clarifying remarks for this sort of problem, since again, it can be quite hard in general! $\endgroup$ – Alex Wertheim Dec 6 '16 at 2:15
  • $\begingroup$ You said you could solve this type of problem using Galois theory, could you give me a link to it,please? $\endgroup$ – J.U.math Dec 6 '16 at 2:18
  • $\begingroup$ @Edgar.W: sure. Here is an example where you can prove that $\sqrt{5} \notin \mathbb{Q}(\sqrt[3]{2})$ using Galois-theoretic methods: math.stackexchange.com/questions/937071/… Of course, there are easier approaches in this case, e.g. degree arguments. $\endgroup$ – Alex Wertheim Dec 6 '16 at 2:22
  • $\begingroup$ I should point out that this is not a great example, since the argument that $[L:\mathbb{Q}] = 6$ is equivalent to the fact that $\sqrt{5} \notin \mathbb{Q}(\sqrt[3]{2})$, but you can see how the line of reasoning works nonetheless. $\endgroup$ – Alex Wertheim Dec 6 '16 at 2:28

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