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Let $\sqrt[3]{2}$ be the real cube root of 2.

(a) Find the minimal polynomials of cube root of $2$ and $i$ over $\mathbb{Q}$.
(b) Find the minimal polynomial of $i$ over $\mathbb{Q}(\sqrt[3]{2})$.
(c) Show that $\mathbb{Q}(i\cdot\sqrt[3]{2}) = \mathbb{Q}(i,\sqrt[3]{2})$.
(d) Find the minimal polynomial of $i\cdot\sqrt[3]{2}$ over $\mathbb{Q}$.

I am lost in this setup, and this is not a homework assignment. It is just for something the teacher didn't cover due to lack of time.

For part a:

I got the answer to be x^3 -2 and x^2 + 1, respectively

For part b:

I got the answer to be x^2 + 1

part c: I am unsure what's going on, and need help. can someone explain it please?

part d: answer is x^6 + 4 = 0, right?

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For a start: what is the minimal polynomial of cube root 2? what is the minimal polynomial of $i$? – Fabian May 3 '11 at 5:18
x^3-2; i^3-2, are there correct? – mary May 3 '11 at 5:26
The first one is correct. Why $i^3-2$? – Dennis Gulko May 3 '11 at 6:11
I posted something about c) ten days ago. If you didn't find it helpful, maybe you could indicate what part(s) of my answer you have trouble with. This is more polite than just editing your question and ignoring (or appearing to ignore) efforts to help you. – Gerry Myerson May 13 '11 at 12:48

Here's a little help with part c). Can you see why it's enough to show that $i\root3\of2$ is in ${\bf Q}(i,\root3\of2)$ and that both $i$ and $\root3\of2$ are in ${\bf Q}(i\root3\of2)$? Can you see that the first of these inclusions is trivial? Can you see why $(i\root3\of2)^4$ is in ${\bf Q}(i\root3\of2)$, and how that helps you to show that both $i$ and $\root3\of2$ are in ${\bf Q}(i\root3\of2)$?

Actually, maybe I should start with this: do you know what ${\bf Q}(\root3\of2)$ means? If you don't know that, that's the place to start.

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Do you understand what a minimum polynomial over $\mathbb{Q}$ is? For example, the minimal polynomial $f$ of $(2)^{1/3}$ over $\mathbb{Q}$ is simply the monic polynomial of least degree with coefficients in $\mathbb{Q}$ such that $f(2^{1/3})=0$. It should be clear that this is just $x^3-2$.

Regarding your comment to Fabian; is $i^3-2$ a polynomial?

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Technically, $i^3-2$ is a polynomial of degree 0, i.e., a constant. Of course, it has neither coefficient in $\mathbb{Q}$ nor has any zeros. – Jiangwei Xue May 3 '11 at 5:38
@Jiangwei - this is true, I guess! In addition to not having coefficients in $\mathbb{Q}$, nor zeroes, it is also not monic! – Juan S May 3 '11 at 5:41

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