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I'm looking for the minimal polynomial of $\alpha = i\frac{\sqrt{3}}{2}+\frac{1}{2}$ in $\mathbb{Q}[x]$. A polynomial with a root $\alpha$ is

$$(2(x-\frac{1}{2}))^4-9.$$

A computer algebra system shows me that the polynomial is irreducible. I looking for a way to compute this by hand in an exam. Is there any way to do so without using any irreduciblity criterions which where not part of my lecture?

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First of all, it is not monic so it is not the minimal polynomial. Second, how, in your lectures, polynomials of degree $>3$ were shown to be irreducible over $\mathbb{Q}$ without irreduciblity criterions ? –  Belgi Sep 2 '12 at 16:22
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btw: wolframalpha.com/input/?i=%282%28x-\frac{1}{2}%29%29^4-9 shows a decomposition into two polynomials of degree $2$ –  Belgi Sep 2 '12 at 16:26
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The polynomial is a difference of squares, so it is immediately obvious that it factors- I point this out as a general technique: it factors as $[4(x-\frac{1}{2})^{2} - 3][4(x-\frac{1}{2})^{2} +3],$ and each of these quadratic factors is in $\mathbb{Q}[x].$ –  Geoff Robinson Sep 2 '12 at 17:03
    
@Joachim: The question may be relevant to others sometime in the future, this is why users lose the ability to delete questions after an answer has been posted. –  Eric Naslund Sep 3 '12 at 1:18

2 Answers 2

up vote 4 down vote accepted

If $x=i\frac{\sqrt3}{2}+\frac{1}{2}\implies 2x-1=i\sqrt3\implies x^2-x+1=0 $ is the minimal polynomial with integral coefficients.

Another way to address this is to put $i\frac{\sqrt3}{2}+\frac{1}{2}=R(\cos y+i\sin y)=Re^{iy}$ where $R$ is a positive real number.

Squaring & adding we get , $R^2=1\implies R=1$

So, $\cos y=\frac{1}{2}$ and $\sin y=\frac{\sqrt3}{2}$ $\implies \tan y=\sqrt3$

As the cosine and sine ratios of $y$ are positive, so $y=\frac{\pi}{3}$

So, $x=i\frac{\sqrt3}{2}+\frac{1}{2}=e^{\frac{i\pi}{3}}$

To rationalize $e^{\frac{ip\pi}{q}}$ where $p,q$ are integers with (p,q)=1, we need to take the q-th power as $e^{ip\pi}=(-1)^p$

So $x^3=-1\implies x^3+1=0$, but clearly,$x≠-1$

So, $\frac{x^3+1}{x+1}=0\implies x^2-x+1=0$

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This polynomial is not monic –  Belgi Sep 2 '12 at 16:25
    
Why monic polynomial is required at all? Am I missing anything? –  lab bhattacharjee Sep 2 '12 at 16:33
    
Due to the definition of minimal polynomial to make it unique. –  joachim Sep 2 '12 at 16:36
    
I'll first need to ask you if you think the minimal polynomial is unique ? (the definition I know for minimal polynomial requires it to be monic, as a side product it is also unique by this definition) –  Belgi Sep 2 '12 at 16:36
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They ARE different as polynomials... –  fretty Sep 2 '12 at 16:47

You can also do this by thinking geometrically - from trigonometry, you might remember that $$e^{i \pi/3} = \frac{1}{2}+ i \frac{\sqrt{3}}{2} .$$ Therefore, this is a zero of $x^3 +1$. A quick check shows that this is not irreducible, since $-1$ is a root. Therefore we can factor it: $x^3 + 1 = (x -1)(x^2 - x + 1)$. Thus $ \frac{1}{2}+ i \frac{\sqrt{3}}{2} $ is a root of $x^2 - x + 1$. The minimal polynomial must have degree $\geq 2$ since $ \frac{1}{2}+ i \frac{\sqrt{3}}{2} \notin \mathbb{Q}. $ Therefore $x^2 -x +1$ must be the minimal polynomial. Alternately, you can check directly that $x^2 -x +1$ is irreducible over $\mathbb{Q}$ since it is degree 2 and its roots are not in $\mathbb{Q}$.

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