# Galois Group of the Polynomial $x^2-2$ ## Problem 230

Let $\Q$ be the field of rational numbers.

(a) Is the polynomial $f(x)=x^2-2$ separable over $\Q$?

(b) Find the Galois group of $f(x)$ over $\Q$. Add to solve later

## Solution.

### (a) The polynomial $f(x)=x^2-2$ is separable over $\Q$

The roots of the polynomial $f(x)$ are $\pm \sqrt{2}$. Since all the roots of $f(x)$ are distinct, $f(x)=x^2-2$ is separable.

### (b) The Galois group of $f(x)$ over $\Q$

The Galois group of the separable polynomial $f(x)=x^2-2$ is the Galois group of the splitting field of $f(x)$ over $\Q$.
Since the roots of $f(x)$ are $\pm \sqrt{2}$, the splitting field of $f(x)$ is $\Q(\sqrt{2})$. Thus, we want to determine the Galois group
$\Gal(\Q(\sqrt{2})/\Q).$ Let $\sigma \in \Gal(\Q(\sqrt{2})/\Q)$. Then the automorphism $\sigma$ permutes the roots of the irreducible polynomial $f(x)=x^2-2$.

Thus $\sigma(\sqrt{2})$ is either $\sqrt{2}$ or $-\sqrt{2}$. Since $\sigma$ fixes the elements of $\Q$, this determines $\sigma$ completely as we have
$\sigma(a+b\sqrt{2})=a+b\sigma(\sqrt{2})=a\pm \sqrt{2}.$

The map $\sqrt{2} \mapsto \sqrt{2}$ is the identity automorphism $1$ of $\Q \sqrt{2}$.
The other map $\sqrt{2} \mapsto -\sqrt{2}$ gives non identity automorphism $\tau$. Therefore, the Galois group $\Gal(\Q(\sqrt{2})/\Q)=\{1, \tau\}$ is a cyclic group of order $2$.

In summary, the Galois group of the polynomial $f(x)=x^2-2$ is isomorphic to a cyclic group of order $2$. Add to solve later

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