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Minimal polynomial of $\frac{\sqrt{5}+1}{2}$

My try:$$x=\frac{\sqrt{5}+1}{2}$$ $$(2x-1)^2=5$$$$x^2-x-1=0$$ i.e of degree 2.Please check.

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    Minimal polynomial over what field? $\mathbb Q$? – 2017-02-08
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    yes minimal polynomial over $\mathbb Q$ – 2017-02-08
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    it is fine to me! – 2017-02-08
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    May I ask what is your definition of a minimal polynomial? – 2017-02-08
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    lowest degree monic polynomial for which $\frac{\sqrt{5}+1}{2}$ is a root over $\mathbb Q$. – 2017-02-08
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    Strictly speaking, you only showed that $x^2-x-1$ is *some* polynomial having $\frac{\sqrt 5+1}2$ as root. But of course it is indeed minimal because any smaller polynomial would have to be linear, i.e., $\frac{\sqrt 5+1}{2}\in\Bbb R$ - which is absurd. – 2017-02-09

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One process of finding the minimal polynomial of $\alpha$ for a simple field extension $K(\alpha):K$ is to first find a polynomial with $\alpha$ as a root, then to prove (or at least argue) that it has the least degree among all such polynomials.

You have successfully found a polynomial, which is $x^2-x-1=0$. Now, how to prove that this polynomial has the least possible degree, i.e. $2$?

When a quadratic polynomials is nominated as the minimal polynomial, we deploy the tactic mentioned by @HagenvonEitzen and attempt a proof by condradiction:

Assume that the minimal polynomial is of degree $1$, then it is of the form $x-\alpha$. In our case, it's $x-\frac{\sqrt{5}+1}{2}$. A minimal polynomial must have coefficients in $K$ ($\mathbb{Q}$ in our case), but $\frac{\sqrt{5}+1}{2}\not\in\mathbb{Q}$ because $\sqrt{5}\not\in\mathbb{Q}$. Therefore, the minimal polynomial of $\frac{\sqrt{5}+1}{2}$ over $\mathbb{Q}$ cannot have degree $1$ and hence must have a degree of $2$ and above, which means that your polynomial $$ x^2-x-1=0 $$ is indeed the minimal polynomial of $x^2-x-1=0$ over $\mathbb{Q}$.