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Find all automorphisms of $\mathbb{Q}(\sqrt[3]{5})$.

How can I solve the above problem ? Please help someone.

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marked as duplicate by JSchlather, Gerry Myerson, Zev Chonoles Mar 3 '13 at 5:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

what have you tried? –  Bombyx mori Mar 3 '13 at 4:02
Every automorphism is perfectly determined by its action on $\sqrt[3]{5}$. –  Mathematician Mar 3 '13 at 4:26
I don't think this is a duplicate of the suggested question. It may be a dup of another one though. Does ring a bell! Ok. It is an abstract duplicate in the sense that the same method leads to an answer, but... :-) –  Jyrki Lahtonen Mar 3 '13 at 5:43

1 Answer 1

An automorphism of $\mathbb{Q}(\sqrt[3]{5})$ is (by definition) a isomorphism $f:\mathbb{Q}(\sqrt[3]{5})\to \mathbb{Q}(\sqrt[3]{5})$.

Note that $$\mathbb{Q}(\sqrt[3]{5})=\{a+b\sqrt[3]{5}+c(\sqrt[3]{5})^2\mid a,b,c\in\mathbb{Q}\}.$$ Thus, if $f$ is an automorphism of $\mathbb{Q}(\sqrt[3]{5})$ and I tell you where $f$ sends $\sqrt[3]{5}$, i.e. if I tell you that $f(\sqrt[3]{5})=\alpha$ for some $\alpha\in\mathbb{Q}(\sqrt[3]{5})$, then you will know what $f$ does to any element of $\mathbb{Q}(\sqrt[3]{5})$ is because $f(t)=t$ for all $t\in\mathbb{Q}$ and $f$ is a ring homomorphism, hence $$f(a+b\sqrt[3]{5}+c(\sqrt[3]{5})^2)=a+bf(\sqrt[3]{5})+cf(\sqrt[3]{5})^2=a+b\alpha+c\alpha^2.$$ Thus, to classify the automorphisms of $\mathbb{Q}(\sqrt[3]{5})$, it suffices to figure out which elements $\alpha\in\mathbb{Q}(\sqrt[3]{5})$ an automorphism $f$ is allowed to send $\sqrt[3]{5}$ to.

Now note that, if $\alpha=f(\sqrt[3]{5})$ for an automorphism $f$, then because $(\sqrt[3]{5})^3-5=0$, we must also have that $$0=f(0)=f((\sqrt[3]{5})^3-5)=f(\sqrt[3]{5})^3-5=\alpha^3-5.$$ Thus, if $f$ is an automorphism of $\mathbb{Q}(\sqrt[3]{5})$ and $f(\sqrt[3]{5})=\alpha\in\mathbb{Q}(\sqrt[3]{5})$, then $\alpha^3=5$.

Clearly, one option is $\alpha=\sqrt[3]{5}$; this corresponds to the identity automorphism (see the computation above). Can you figure out if there are any other $\alpha\in\mathbb{Q}(\sqrt[3]{5})$ that work?

If I have put real effort into figuring it out from the above, but you are still stuck, you can mouse over the gray area below for a further hint.

Hint: what are the $\alpha\in\mathbb{C}$ with the property that $\alpha^3=5$? Are they elements of $\mathbb{Q}(\sqrt[3]{5})$?

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is the answer will be 3? –  user59908 Mar 3 '13 at 4:54
I don't understand what you mean. Are you proposing that $\alpha=3$ works? –  Zev Chonoles Mar 3 '13 at 4:56
the number of automorphism is 3.am I right? –  user59908 Mar 3 '13 at 5:24
No, that's not correct. Are the other two cube roots of 5 elements of $\mathbb{Q}(\sqrt[3]{5})$? (Hint: they are not real numbers) –  Zev Chonoles Mar 3 '13 at 5:28
+1 to Zev. @user59908: Plot $x^3-5$. How many times does its graph cross the $x$-axis? –  Jyrki Lahtonen Mar 3 '13 at 5:40

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