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If $\alpha$ is a root of $x^3-3x+1$ then $f: \mathbb{Q}(\alpha)\to \mathbb{Q}(\alpha)$ such that $f(1)=1$ and $f(\alpha)=\alpha^2-2$ is a automorphism.

Assuming that is a homomorphism, then is fair easy to check that is an a biyection. Since for any $a+b\alpha+c\alpha^2 \in \mathbb{Q}(\alpha)$, $$f\left(a+2(c-b)+4b+\alpha(c-b)-b\alpha^2\right)=a+b\alpha+c\alpha^2$$ And $\mathbb{Q}(\alpha)/\operatorname{Ker}(f)\cong \mathbb{Q}(\alpha)$, then f is a biyection.

The problem is that I don't know how to prove that it really is a homomorphism. Also, i would also like to know alternative or smart ways to show this

Theincredibleidiot
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Nicolas Rodriguez
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    Check whether $\alpha^2-2$ is also a root of $f$. – Arturo Magidin Sep 02 '24 at 02:09
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    I already do it, $\alpha^2-2$ and $2-\alpha-\alpha^2$ are the 3 roots with $\alpha$ and all of them are in $\mathbb{Q}(\alpha)$. But idk how to actually write $f(a+b\alpha+c\alpha^2)=f(a)+f(b)f(\alpha)+f(c)f(\alpha^2)$ – Nicolas Rodriguez Sep 02 '24 at 02:14
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    Well, then, you are done. This is a standard theorem of field extensions: If $p(x)$ is a polynomial that is irreducible over $\mathbb{F}$, and $\beta$ and $\gamma$ are two roots of $p(x)$, then there is an isomorphism $\mathbb{F}[\alpha]\to\mathbb{F}[\beta]$ that fixes $\mathbb{F}$ and sends $\alpha$ to $\beta$. That is what you are doing, so all you need is note that since $\mathbb{Q}[\alpha^2-2]\subseteq\mathbb{Q}[\alpha]$ and they both have degree $3$ over $\mathbb{Q}$, they are actually equal. – Arturo Magidin Sep 02 '24 at 02:18
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    If you wish to be very explicit in the real numbers, the roots are $ 2 \cos (2 \pi / 9), 2 \cos (4 \pi / 9), 2 \cos (8 \pi / 9) $ – Cranium Clamp Sep 02 '24 at 02:38
  • I trust u, but that is what actually have to prove. Namely a morphism that fix all the constants terms and "permutes" the roots is biyective. – Nicolas Rodriguez Sep 02 '24 at 02:47
  • The thing is that im on a algebra class and this exercise is a essay. The next topic is Galois theory and im pretty sure that this is a theorem, Thou, i cannot yet used it – Nicolas Rodriguez Sep 02 '24 at 02:52
  • First, any map will fix the rationals necessarily, so $f(a)=a$, $f(b)=b$, and $f(c)=c$. Then all you need to do is do the calculation. – Arturo Magidin Sep 02 '24 at 03:05
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    The definition of $f$ is incomplete. The definition must give a way to evaluate $f(\beta) $ where $\beta$ is any arbitrary element of $\mathbb{Q} (\alpha) $. Just giving values of $f$ at $1,\alpha$ doesn't help. Or perhaps the question is already assuming that $f$ is a homomorphism. – Paramanand Singh Sep 02 '24 at 03:07
  • @ArturoMagidin i will give a shot, but before hand i can see that $f|_{\mathbb{Q}}$ is the identity map. – Nicolas Rodriguez Sep 02 '24 at 03:14
  • @ParamanandSingh i would ask that to my teacher. thanks. Btw that is my problem, i dont have a explicit form of $f$ – Nicolas Rodriguez Sep 02 '24 at 03:14
  • If $f(1)=1$, then by induction $f(n)=n$ for all positive integers. SInce $f(n)+f(-n)=f(0)=0$, then $f(a)=a$ for all integers. Since $1 = f(1) = f(p/p) = f(p)f(1/p) = pf(1/p)$, then $f(1/p)=1/p$ for all integers $p$. Then $f(p/q) = f(p)f(1/q) = p/q$. – Arturo Magidin Sep 02 '24 at 03:21
  • This polynomial recurs in exercises on field theory - for a reason! See also here. I answered that one, so I'm hesitant to vote this as a duplicate, but anyway. Check out the other threads linked to from there :-) – Jyrki Lahtonen Sep 02 '24 at 03:27
  • @JyrkiLahtonen U could and u right, this exercise come from my algebra class, but the spirit of the question is not the same. – Nicolas Rodriguez Sep 02 '24 at 23:20

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Note that $\{1,\alpha,\alpha^2\}$ is a basis of $\mathbb Q(\alpha)$. If you announce the values of $f$ at these points then it extends to a homomorphism. Here the $f:\mathbb Q(\alpha)\to \mathbb Q(\alpha)$ is defined by $$a+b\alpha+c\alpha^2\mapsto a+b(\alpha^2-2)+c(\alpha^2-2)^2=(a-2b+4c)-c\alpha+(b-c)\alpha^2,$$ where the fact $\alpha^3=3\alpha-1$ has been used.

Theincredibleidiot
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  • Thanks, so i can build the homomorphism given that $f(1)=1$ and $f(\alpha)=\alpha^2-2$ and that is all of it? – Nicolas Rodriguez Sep 02 '24 at 03:17
  • Yes. You are right. – Theincredibleidiot Sep 02 '24 at 03:22
  • If you announce the values of $f$ at elements of a basis, then it extends to a linear transformation. It is anything but given that you get a homomorphism that way. In fact, to get a homomorphism, you have to define $f(1)=1$. In this case there are exactly three possible choice for $f(\alpha)$. Namely $\alpha$, $\alpha^2-2$ and $(\alpha^2-2)^2-2$. The last alternative simplifies to $2-\alpha-\alpha^2$. See Arturo Magidin's comments under the question. – Jyrki Lahtonen Sep 02 '24 at 03:36
  • @JyrkiLahtonen Here I think it is assumed that $f(x^n)=f(x)^n$ and other similar things. So a field homomorphism is obtained. – Theincredibleidiot Sep 02 '24 at 03:42
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    Only when $\alpha$ is mapped to a zero of the same polynomial $x^3-3x+1$. Otherwise it won't be a homomorphism. If $\alpha^3-3\alpha+1=0$, then we also need $$0=f(0)=f(\alpha^3-3\alpha+1)=f(\alpha^3)-f(3)f(\alpha)+f(1)=f(\alpha)^3-3f(\alpha)+1.$$ – Jyrki Lahtonen Sep 02 '24 at 04:10
  • Yes, that is correct. – Theincredibleidiot Sep 02 '24 at 04:38
  • Thx, that is what i needed. – Nicolas Rodriguez Sep 02 '24 at 23:18