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 $\Q$.
Sponsored Links
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
Sponsored Links
More from my site
Galois Extension $\Q(\sqrt{2+\sqrt{2}})$ of Degree 4 with Cyclic Group
Show that $\Q(\sqrt{2+\sqrt{2}})$ is a cyclic quartic field, that is, it is a Galois extension of degree $4$ with cyclic Galois group.
Proof.
Put $\alpha=\sqrt{2+\sqrt{2}}$. Then we have $\alpha^2=2+\sqrt{2}$. Taking square of $\alpha^2-2=\sqrt{2}$, we obtain […]- Galois Group of the Polynomial $x^p-2$.
Let $p \in \Z$ be a prime number.
Then describe the elements of the Galois group of the polynomial $x^p-2$.
Solution.
The roots of the polynomial $x^p-2$ are
\[ \sqrt[p]{2}\zeta^k, k=0,1, \dots, p-1\]
where $\sqrt[p]{2}$ is a real $p$-th root of $2$ and $\zeta$ […]
Determine the Splitting Field of the Polynomial $x^4+x^2+1$ over $\Q$
Determine the splitting field and its degree over $\Q$ of the polynomial
\[x^4+x^2+1.\]
Hint.
The polynomial $x^4+x^2+1$ is not irreducible over $\Q$.
Proof.
Note that we can factor the polynomial as […]- $x^3-\sqrt{2}$ is Irreducible Over the Field $\Q(\sqrt{2})$
Show that the polynomial $x^3-\sqrt{2}$ is irreducible over the field $\Q(\sqrt{2})$.
Hint.
Consider the field extensions $\Q(\sqrt{2})$ and $\Q(\sqrt[6]{2})$.
Proof.
Let $\sqrt[6]{2}$ denote the positive real $6$-th root of of $2$.
Then since $x^6-2$ is […]
- Automorphism Group of $\Q(\sqrt[3]{2})$ Over $\Q$.
Determine the automorphism group of $\Q(\sqrt[3]{2})$ over $\Q$.
Proof.
Let $\sigma \in \Aut(\Q(\sqrt[3]{2}/\Q)$ be an automorphism of $\Q(\sqrt[3]{2})$ over $\Q$.
Then $\sigma$ is determined by the value $\sigma(\sqrt[3]{2})$ since any element $\alpha$ of $\Q(\sqrt[3]{2})$ […]
- The Polynomial $x^p-2$ is Irreducible Over the Cyclotomic Field of $p$-th Root of Unity
Prove that the polynomial $x^p-2$ for a prime number $p$ is irreducible over the field $\Q(\zeta_p)$, where $\zeta_p$ is a primitive $p$th root of unity.
Hint.
Consider the field extension $\Q(\sqrt[p]{2}, \zeta)$, where $\zeta$ is a primitive $p$-th root of […]
- Two Quadratic Fields $\Q(\sqrt{2})$ and $\Q(\sqrt{3})$ are Not Isomorphic
Prove that the quadratic fields $\Q(\sqrt{2})$ and $\Q(\sqrt{3})$ are not isomorphic.
Hint.
Note that any homomorphism between fields over $\Q$ fixes $\Q$ pointwise.
Proof.
Assume that there is an isomorphism $\phi:\Q(\sqrt{2}) \to \Q(\sqrt{3})$.
Let […]
- Cubic Polynomial $x^3-2$ is Irreducible Over the Field $\Q(i)$
Prove that the cubic polynomial $x^3-2$ is irreducible over the field $\Q(i)$.
Proof.
Note that the polynomial $x^3-2$ is irreducible over $\Q$ by Eisenstein's criterion (with prime $p=2$).
This implies that if $\alpha$ is any root of $x^3-2$, then the […]
