Galois Theory - field extensions and irreducible polynomials 

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*Let $n\in\mathbb{N}$. Find the extension degree and a basis of $\mathbb{Q}(\sqrt[n]{3})$ over $\mathbb{Q}$.





*Prove that $\sqrt[3]{3}\not\in\mathbb{Q}(\sqrt[4]{3})$.





*Prove that $x^3-3$ is irreducible over $\mathbb{Q}(\sqrt[4]{3})$.


Attempt: 1. I claim that the extension degree is $n$, with following basis: $\{1,\sqrt{3},\sqrt[3]{3},\cdots,\sqrt[n]{3}\}$. This observation comes from looking at small cases, and the argument generalises. I'm not sure how to make it rigorous though.


*We need only show that if $a,b\in\mathbb{Q}$ then $\sqrt[3]{3}=a+b\sqrt[4]{3}$ has no solutions. But this is clear - due to our basis in part 1 we have that $\sqrt[i]{3}$ and $\sqrt[j]{3}$ are linearly independent over $\mathbb{Q}$ for $i\ne j$.


*Define $f(x)=x^3-3$. Note that $f(\sqrt[3]{3})=0$ and is hence a root. This then factors into a linear and a quadratic. The linear term is not in the field as we proved in the last part. I don't know whether this solves the problem though - I suppose we should also consider the quadratic term.
 A: In general, for this sort of problem, one should not try to appeal to the basic definitions or brute force arguments like you're doing for part 1 and 2. Field theory is rich, and there are lots of more general theorems at our disposal that are more appropriate for solving problems like these.

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*Clearly, $\sqrt[n]{3}$ is a root of $f(x)=x^n-3$ which is irreducible by Eisenstein's criterion, so $[\mathbb{Q}(\sqrt[n]{3})/\mathbb{Q}]=\deg(f)=n$.


*Suppose that $\sqrt[3]{3}\in\mathbb{Q}(\sqrt[4]{3})$ then $\mathbb{Q}(\sqrt[3]{3})$ is an intermediate field of $\mathbb{Q}(\sqrt[4]{3})/\mathbb{Q}$, so
$$[\mathbb{Q}(\sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{3})]=\frac{[\mathbb{Q}(\sqrt[4]{3}):\mathbb{Q}]}{[\mathbb{Q}(\sqrt[3]{3}):\mathbb{Q}]}=\frac{4}{3}$$
which is non-integer; a contradiction. Therefore $\sqrt[3]{3}\not\in\mathbb{Q}(\sqrt[4]{3})$.


*Note that a polynomial of degree $3$ in $k[x]$ is irreducible over a field $k$ iff it has no roots in $k$ (precisely because of the factoring you mentioned). Observe that $f(x)$ has three roots $\sqrt[3]{3},e^{i\pi 2/3}\sqrt[3]{3},e^{i\pi 4/3}\sqrt[3]{3}$. We know that $\sqrt[3]{3}\not\in\mathbb{Q}(\sqrt[4]{3})$, and since the other two roots are complex, then they are not in $\mathbb{Q}(\sqrt[3]{4})$ either. Therefore, $f(x)$ is irreducible over $\mathbb{Q}(\sqrt[4]{3})$.
