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Find the minimal polynomial over $\mathbb Q$ and $\mathbb R$ for ...$\sqrt[3]{3}$, $1- i\sqrt{3}$, $2 + i$, $i\sqrt[3]{3}$

Sorry for my sqrt formulas .. I'm new here, hope to learn really fast to write a correct question in math symbols.

I have to find both minimal polynomial over $\mathbb Q$, as well over $\mathbb R$ ... It's kind of homework ... Any good explanations will be nice and very appreciated. Thanks

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Welcome on math.SE. To write a question you can use standard $\LaTeX$ markup, and you should mark a homework question with (homework). – martini May 22 '13 at 10:19
For the $\mathbb{R}$ case, notice that if $f$ is the minimal polynomial of a non-real complex number $c$, then also $\bar{c}$ (conjugate of $c$) is a root of $f$. – egreg May 22 '13 at 10:26
up vote 1 down vote accepted

I'll do the first one for you as hint.

Let $\alpha = \sqrt[3]{3}$

Let's get rid of that pesky radical by cubing both sides:

$\alpha^3 = 3$

Then move 3 over to the other side:

$\alpha^3 - 3 = 0$

Let $f = x^3 - 3$. Now f is irreducible in $\mathbb Q[x]$ by Eisenstein's Criterion (p=3) and Gauss' lemma. So the minimal polynomial for $\sqrt[3]{3}$ over $\mathbb Q$ is $m_\sqrt[3]{3} = x^3 - 3$.

The rest follow a very similar procedure.

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Thanks! But f is not irreducible in R[x], right ? – Florin M. May 22 '13 at 10:28
@MFG Flay: That is correct. That case is actually quite trivial. – Islands May 22 '13 at 10:29
THank you for your response. I appreciate it. But I think I will need some other explanation for the R case – Florin M. May 22 '13 at 10:31
For $\mathbb R$ note that $\sqrt[3]3 \in \mathbb R$, hence $X -\sqrt[3]3 \in \mathbb R[X]$ ... and linear polynomials are irreducible. – martini May 22 '13 at 10:33
@MFG Flay: You really should try the others yourself using the method in my post. If you get stuck, post what you've done and I'll be glad to help. – Islands May 22 '13 at 10:38

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