[Math] Minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$

abstract-algebraextension-fieldminimal-polynomials

I had an example in the book given as follows:

Find the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$ .

Solution: $~~~~~$$(\sqrt{2}+\sqrt{3})^2=5+2 \sqrt6$

$(\sqrt{2}+\sqrt{3})^4=49+20 \sqrt6$

Then $(\sqrt{2}+\sqrt{3})^4-10(\sqrt{2}+\sqrt{3})^2+1=0.$

Thus $a=\sqrt{2}+\sqrt{3}$ satisfies $f(x)=x^4-10x^2+1$ over $\mathbb Q$.

Let $p(x)$ be the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbb Q$.Then $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ are also roots of $p(x).$
So degree of $p(x)$ is atleast $4$.

But $f(a)=0 $ and $f(x) \in \mathbb Q[x] \implies p(x)$ divides $f(x)$ .

So $f(x)$ is minimal polynomial of $\sqrt{2}+\sqrt{3}.$

But I can't get the step how are $(\sqrt{2}-\sqrt{3}),(-\sqrt{2}+\sqrt{3}),(-\sqrt{2}-\sqrt{3})$ also roots of $p(x).$

Kindly help with this.

Best Answer

That fact comes from Galois Theory. Under the assumption that $K/F$ is Galois, the (monic) minimal polynomial $f$ of an element $\alpha \in K$ (over $F$) is of the form $$f(x) = \prod_{\sigma \in \text{Gal}(K/F)} (x - \sigma \alpha).$$ This implies that $\sigma \alpha$ are all roots of the minimal polynomial for $\alpha$.

In your case, let $K = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ and $F = \mathbb{Q}$. Then $K/F$ is Galois and there are 4 automorphisms in $\text{Gal}(K/F)$: namely

  • identity on $K$
  • the one that sends $\sqrt{2} \mapsto -\sqrt{2}$ keeping $\sqrt{3}$ fixed: this automorphism sends $\sqrt{2} + \sqrt{3}$ to $-\sqrt{2} + \sqrt{3}$; hence $-\sqrt{2} + \sqrt{3}$ must be a root of $p(x)$.
  • the one that keeps $\sqrt{2}$ fixed and sends $\sqrt{3} \mapsto -\sqrt{3}$: like above, we deduce $\sqrt{2} - \sqrt{3}$ is root of minimal polynomial $p(x)$
  • the one that negate both roots: finally $- \sqrt{2} - \sqrt{3}$ must also be a root.
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