Automorphism Group of $\Q(\sqrt[3]{2})$ Over $\Q$.

Field theory problems and solution in abstract algebra

Problem 97

Determine the automorphism group of $\Q(\sqrt[3]{2})$ over $\Q$.

LoadingAdd to solve later

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 $\Q(\sqrt[3]{2})$ can be written as $\alpha=a+b\sqrt[3]{2}+c\sqrt[3]{2}^2$ for some $a,b,c \in \Q$ and
\begin{align*}
\sigma(\alpha) &=\sigma(a+b\sqrt[3]{2}+c\sqrt[3]{2}^2)\\
&=\sigma(a)+\sigma(b)\sigma(\sqrt[3]{2})+\sigma(c)\sigma(\sqrt[3]{2})^2\\
&=a+b\sigma(\sqrt[3]{2})+c\sigma(\sqrt[3]{2})^2
\end{align*}
since $\sigma$ fixes the elements in $\Q$.

Note that $\sigma(\sqrt[3]{2})$ is a root of the minimal polynomial $x^3-2$ of $\sqrt[3]{2}$.
But the roots of $x^3-2$ are not real except $\sqrt[3]{2}$, hence not in $\Q(\sqrt[3]{2})$.

Thus we must have $\sigma(\sqrt[3]{2})=\sqrt[3]{2}$. Hence $\sigma$ is trivial.
In conclusion, we have
\[\Aut(\Q(\sqrt[3]{2})/Q)=\{1\}.\]


LoadingAdd to solve later

Sponsored Links

More from my site

  • Galois Group of the Polynomial $x^2-2$Galois Group of the Polynomial $x^2-2$ 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 $\Q$.   Solution. (a) The polynomial $f(x)=x^2-2$ is separable over $\Q$ The roots of the polynomial $f(x)$ are $\pm […]
  • Galois Extension $\Q(\sqrt{2+\sqrt{2}})$ of Degree 4 with Cyclic GroupGalois 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 […]
  • Abelian Normal subgroup, Quotient Group, and Automorphism GroupAbelian Normal subgroup, Quotient Group, and Automorphism Group Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$. Let $\Aut(N)$ be the group of automorphisms of $G$. Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime. Then prove that $N$ is contained in the center of […]
  • Galois Group of the Polynomial  $x^p-2$.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 $\zeta$ […]
  • Degree of an Irreducible Factor of a Composition of PolynomialsDegree of an Irreducible Factor of a Composition of Polynomials Let $f(x)$ be an irreducible polynomial of degree $n$ over a field $F$. Let $g(x)$ be any polynomial in $F[x]$. Show that the degree of each irreducible factor of the composite polynomial $f(g(x))$ is divisible by $n$.   Hint. Use the following fact. Let $h(x)$ is an […]
  • $x^3-\sqrt{2}$ is Irreducible Over the Field $\Q(\sqrt{2})$$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 […]
  • Equation $x_1^2+\cdots +x_k^2=-1$ Doesn’t Have a Solution in Number Field $\Q(\sqrt[3]{2}e^{2\pi i/3})$Equation $x_1^2+\cdots +x_k^2=-1$ Doesn’t Have a Solution in Number Field $\Q(\sqrt[3]{2}e^{2\pi i/3})$ Let $\alpha= \sqrt[3]{2}e^{2\pi i/3}$. Prove that $x_1^2+\cdots +x_k^2=-1$ has no solutions with all $x_i\in \Q(\alpha)$ and $k\geq 1$.   Proof. Note that $\alpha= \sqrt[3]{2}e^{2\pi i/3}$ is a root of the polynomial $x^3-2$. The polynomial $x^3-2$ is […]
  • Application of Field Extension to Linear CombinationApplication of Field Extension to Linear Combination Consider the cubic polynomial $f(x)=x^3-x+1$ in $\Q[x]$. Let $\alpha$ be any real root of $f(x)$. Then prove that $\sqrt{2}$ can not be written as a linear combination of $1, \alpha, \alpha^2$ with coefficients in $\Q$.   Proof. We first prove that the polynomial […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Field Theory
Field theory problems and solution in abstract algebra
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 \[x^4+x^2+1.\]

Close