Find the minimal polynomial for a = π^2 over Q(π^3)

**lteece89** · March 26th 2012, 10:11 AM

If it were in Q then the polynomial would be p(x) = x - π^2, but I don't know if being over Q(π^3) changes that. I think that when you compare the fields you get degree 1, but again, I'm not completely sure.

**Sylvia104** · March 26th 2012, 10:33 AM

If the extension is $\mathbb Q(\pi)/\mathbb Q\left(\pi^3\right)$ the minimal polynomial is $x^3-\pi^6.$

**lteece89** · March 26th 2012, 11:54 AM

Is this because the minimal polynomial needs to be degree three?

**Sylvia104** · March 26th 2012, 01:40 PM

The minimal polynomial here is the monic polynomial $f(x)$ of minimal degree with coefficients in $\mathbb Q\left(\pi^3\right)$ such that $f\left(\pi^2\right)=0.$ If $f(x)$ has degree $1$ or $2,$ $f\left(\pi^2\right)$ is of the form $a_0+\pi^2$ or $a_0+a_1\pi^2+\pi^4,$ where $a_0,a_1\in\mathbb Q\left(\pi^3\right);$ clearly these cannot be $0.$

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