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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!

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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"?

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  • $\begingroup$ So the splitting field of f(t) should be the smallest extension of Q which contains all of the solutions to f(t). Should I form a tower of fields of the 3 solutions and examine them one by one? $\endgroup$ – Luke8ball Apr 12 '13 at 18:17
  • $\begingroup$ You could do that, for sure; that would probably work well in this case. Or you might find some way of figuring out beforehand what the splitting field is. $\endgroup$ – Greg Martin Apr 12 '13 at 21:27
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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 $$ [F:\mathbb{Q}]=[F:\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]=2\cdot 3=6. $$

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