Let $K=\mathbb{Q}(i), L=K(\sqrt[3]{3+i}), M=K(\sqrt{\pi}/2)$. Find all $K$-automorphisms of $L$ and $M$.

Question

Let $K=\mathbb{Q}(i), L=K(\sqrt[3]{3+i}), M=K(\sqrt{\pi}/2)$. Find all $K$-automorphisms of $L$ and $M$.

Beside the definitions of what an automorphism is and an idea about how the elements of $L$ and $M$ kind of look like, I am completely lost.

The practical applications of automorphisms which are so important in the chapters which I am doing right now are extremely confusing so I would be really happy for some help.

Thanks

Answer 1

I apport an extend solution on the first extension. I will try to be the clearer, and extense on my thoughts as I can be.

• $\boxed{L=K(\sqrt[3]{3+i})}$ The $K$-automorphisms $\sigma: K(\sqrt[3]{3+i})\to K(\sqrt[3]{3+i})$ are group automorphisms that ''fix'' $K$ (i.e., $\sigma(x)=x, \forall x\in K$). Firstly, let's study a little bit the extension $K(\sqrt[3]{3+i})/K$; $x^3-(3+i)$ is the minimal polynomial of $\sqrt[3]{3+i}$, as it is in $K[x]$, is monic, it gets to be $0$ when we evaluate in $x=\sqrt[3]{3+i}$, and is irreducible in $K[x]$; this last fact is easily seen by assuming the opposite thing: if it is reducible, there would exist $a,b,c\in K$ such that $x^3-(3+i)=(x^2+ax+b)(x+c)$; multiplying we arrive to some equalities among the coeficients: $a=0$, $b+ac=0$, $bc=-(3+i)$; but it is easy to see that if $a=0$, $b=0$, and then $bc=0\neq -(3+i)$, so we can conclude that the polynomial is irreducible. So, taking all that we have seen in account, we can conclude that the degree of the extension is $3$ (the degree of the minimal polynomial previously seen), i.e $[K(\sqrt[3]{3+i}):K]=3$; moreover, one basis of this extension (as a $K$-vector space), is $\{1, \sqrt[3]{3+i}, (\sqrt[3]{3+i})^{2}\}$. Let's return to our aim: calculating the $K$-automorphisms; let's take a general element on the extension, expressed in terms of the basis previously calculated: $x=a+b\sqrt[3]{3+i}+c(\sqrt[3]{3+i})^{2}$, with $a,b,c\in K$; if we take a general $K$-automorphism, $\sigma$, we can apply this automorphism in $x$; moreover, taking on account that $K$ is fixed by any $K$-automorphism, and that, as $\sigma$ is an automorphism, $\forall x,y, \sigma(x\cdot y)=\sigma(x)\sigma(y)$ and $\forall x,y, \sigma(x+y)=\sigma(x)+\sigma(y)$: $$ \sigma(x)=\sigma(a+b\sqrt[3]{3+i}+c(\sqrt[3]{3+i})^{2})=\sigma(a)+\sigma(b)\sigma(\sqrt[3]{3+i})+\sigma(c)(\sigma(\sqrt[3]{3+i}))^{2}=$$ $$=a+b\sigma(\sqrt[3]{3+i})+c(\sigma(\sqrt[3]{3+i}))^{2}$$ So, we can conclude that any $K$-automorphism can be uniquely defined by $\sigma(\sqrt[3]{3+i})$; but this is not it: as $(\sqrt[3]{3+i})^{3}-(3+i)=0$, $\sigma((\sqrt[3]{3+i})^{3}-(3+i))=\sigma(0)=0$, so $(\sigma(\sqrt[3]{3+i}))^{3}-(3+i)=0$; that last thing mean that $\sigma(\sqrt[3]{3+i})$ is going to be a root of the minimal polynomial of $\sqrt[3]{3+i}$. So, the maximal number of $K$-automorphisms that we are going to be having is going to be $3$, and that number is only going to be achieved if every root of the minimal polynomial of $\sqrt[3]{3+i}$ is in $K(\sqrt[3]{3+i})$; let's compute the roots of the polynomial $x^3-(3+i)$; if we divide that polynomial by $x-(\sqrt[3]{3+i})$, we obtain $x^2+x\sqrt[3]{3+i} +\sqrt[3]{(3+i)^{2}}$; from this polynomial of degree $2$, let's compute (by the well-known formula for solving cuadratic equations), the other two roots of the minimal polynomial. With this, we obtain $x=-\frac{\sqrt[3]{3+i}}{2}\pm \frac{\sqrt{9+3i}}{2}i$; so, what we have to find is if $-\frac{\sqrt[3]{3+i}}{2}\pm \frac{\sqrt{9+3i}}{2}i\in K(\sqrt[3]{3+i})$,i.e, if $-\frac{\sqrt[3]{3+i}}{2}\pm \frac{\sqrt{9+3i}}{2}i=a+b\sqrt[3]{3+i}+c(\sqrt[3]{3+i})^{2}$, with $a,b,c\in K$; it's quite evident that this is not possible (computation by just calculating is quite tedious, but it is possible to do it; the main problem is that $\sqrt{}$, which is not possible to put in terms of the basis). So, the other two roots of the minimal polynomial are no possible values for $\sigma(\sqrt[3]{3+i})$, and we conclude that the only possible value of $\sigma(\sqrt[3]{3+i})$ is $\sqrt[3]{3+i}$, and we conclude that the only $K$-automorphism of $K(\sqrt[3]{3+i})$ is the identity.

Answer 2

For the extension $L/K$, we first show $x^3-(3+i)$ is irreducible over $K=\mathbb Q(i)$. Since $\mathbb Z[i]$ is integrally closed, we only need to show $x^3-(3+i)$ has no root in $\mathbb Z[i]$, and indeed if there is a solution $x^3=3+i$, then $x^3\overline{x^3} = (x \overline x)^3 = (3+i)(3-i)=10$, but $10$ is not a cubic of any integer. In particular, we conclude that $[L:K]=\deg (x^3-(3+i))=3$.

For any $\sigma\in \text{Auto}(L/K)$, denote $a= \sqrt[3]{3+i}$, we have $\sigma(a)$ must also satisfy the polynomial $x^3-(3+i)$. The roots of this polynomial are obviously $a$ (more precisely a fixed choice of $\sqrt[3]{3+i}$ in $\mathbb C$), $\omega a$, $\omega^2 a$ where $\omega=e^{2\pi i/3}$ and $\omega^2$ are both 3rd roots of unity. Therefore if $\sigma(a)\not=a$, then $L$ contains $\sigma(a)/a$ which is a (primitive) third root of unity. But $[K[\omega]:K]=2\not\mid [L:K]=3$, hence $K[\omega]$ cannot be a subextension of $L/K$. This proves that $\sigma(a)=a$, hence $\sigma = \text{id}_L$.

For the extension $M/K$, since $\pi$ and therefore $\sqrt{\pi}/2$ are transcendental over $K$, we only need to determine $\text{Aut} (K(x)/K)$. It is a standard result that every automorphism of $K(x)/K$ is given by a linear fractional transformation $x\mapsto\frac{ax+b}{cx+d}$. Details can be found in e.g. Galois group of $K(X)/K$