Field Extension With Cube Root of 7 I have the following problem that I am stuck on:
Consider the element $a=\sqrt[3]{7}$ of $\mathbf{R}$. Show that this element is algebraic over $\mathbf{Q}$ and find its minimal polynomial. Also, find the degree of the extension $[\mathbf{Q}\left(\sqrt[3]{7}\right):\mathbf{Q}]$ and find a basis of $\mathbf{Q}\left(\sqrt[3]{7}\right)$ over $\mathbf{Q}$.
My thoughts so far: I think that the minimal polynomial is $f(x)=x^{3}-7$. Is this correct? If so, how to I prove this is the minimal polynomial? Also, I think that the basis should be $\{1,\sqrt[3]{7},\sqrt[3]{49}\}$ based on some other examples I have seen. Is this correct? If so, could someone explain why this is the correct basis? If not, what is the basis?
Thanks in advance for any help!
 A: Of course $f(x) = x^3-7$ has $\sqrt[3]{7}$ as a root, so you just need to show that $f$ is irreducible to show that it is indeed the minimal polynomial of $\sqrt[3]{7}$.  What irreducibility tests do you know?  
Also, your basis is correct.  Suppose we have an algebraic extension $\mathbb{Q}(\alpha)/ \mathbb{Q}$.  If the degree of the extension is $n$, then $\{\alpha^k\}_{k=0}^{n-1}$ serves as a basis.
Proof:  Suppose we could write $\displaystyle \sum_{k=0}^{n-1} c_k \alpha^k = 0$ for $c_k \in \mathbb{Q}$ not all zero.  Then $\displaystyle \sum_{k=0}^{n-1} c_kx^k$ is a polynomial for which $\alpha$ is a root: a contradiction because the minimal polynomial for $\alpha$ must have degree $n = [ \mathbb{Q}(\alpha) : \mathbb{Q}]$.  Furthermore, that we must have $c_k = 0$ for all $0 \leq k \leq n-1$ demonstrates the linear independence of the set $\{\alpha^k\}_{k=0}^{n-1}$.  The cardinality of that linearly independent set is equal to the degree of the extension, so we can rest assured that it is indeed a basis for the extension**.

**Just to drive home fundamentally important ideas: recall that  the degree of an extension $K/F$ is defined to be the dimension of $K$ when $K$ is viewed as a vector space over $F$.  The dimension of a vector space over a field is defined to be the cardinality of a given basis for the vector space.  And it is a theorem that all possible bases of a vector space have equal cardinality, and that any linearly independent set of vectors with that cardinality is necessarily a basis.
A: Since $a$ is clearly a root of $f$, it's enough to show that $f$ is irreducible over $\mathbb{Q}$. Since $f$ is cubic, it suffices to show that $f$ has no roots in $\mathbb{Q}$.
