# Help Determining Degree of a Field Extension

Question: Determine the degree of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$, where $\alpha^3=2$. Determine the degree of the splitting field of $f(t) = t^3 - 2$ over $\mathbb{Q}$.

Is there a difference between these two questions? To answer the first part, I attempted to say that $\alpha$ is algebraic over $\mathbb{Q}$ since it is a solution of $f(t)$, and since it is irreducible over $\mathbb{Q}$ of degree $3$, then the degree of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$ should be $3$. Can anyone use simple terms to explain how to answer these types of questions? I feel lost!

You're right about the first part: the degree of $\mathbb Q(\alpha)$ over $\mathbb Q$ does equal 3. (For another explanation: every element in $\mathbb Q(\alpha)$ can be written uniquely in the form $x+y\alpha+z\alpha^2$ with $x,y,z\in\mathbb Q$, since $\alpha^3=2$; therefore $\mathbb Q(\alpha)$ is a $3$-dimensional vector space over $\mathbb Q$, as suspected.)
As for the second part: what is the splitting field of $f(t)$? I suspect that once you identify that mathematical object, you'll be able to see that the two questions are indeed different. What is the general definition of "splitting field"?
First you are right in saying that $\mathbb{Q}(a)$ is of degree $3$!
Let $F$ be the splitting field of $f(t)$. Let $\zeta=r^{2\pi i/3}=\frac{-1+\sqrt{3}i}{\sqrt{2}}$ be a primitive $3$-rd root of unity and $\alpha=\sqrt[3]{2}$. Then the roots of $f(t)$ are $\alpha, \zeta \alpha, \zeta^2\alpha$; so we have $$F=\mathbb{Q}(\alpha,\zeta\alpha,\zeta^2\alpha)=\mathbb{Q}(\alpha,\zeta).$$
Now $\zeta\not \in \mathbb{Q}(\alpha)$, since $\mathbb{Q}(\alpha)\subseteq \mathbb{R}$ and $\zeta\not\in \mathbb{R}$. Therefore $$[\mathbb{Q}(\alpha,\zeta):\mathbb{Q}(\alpha)]=[F:\mathbb{Q}(\alpha)]\geq 2.$$ On the other hand, as $t^3-1=(t-1)(t^2+t+1)$, it follows that $$[\mathbb{Q}(\alpha,\zeta):\mathbb{Q}(\alpha)]\leq 2,$$ so equality holds and we have $$[F:\mathbb{Q}]=[F:\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]=2\cdot 3=6.$$