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Consider $\mathbb{F}_3(\alpha)$ where $\alpha^3 - \alpha +1 = 0$ and $\mathbb{F}_3(\beta)$ where $\beta^3 - \beta^2 +1 =0$.

I know these two fields are isomorphic but I have difficulty buliding an isomorphism between them.

I know I have to determine where $\alpha$ is mapped to under the isomorphism map but I can't figure it out.

Any help is much appreciated.

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  • $\begingroup$ Well, clearly $\alpha$ has to be mapped to something which is a root of $x^3-x+1=0$. $\endgroup$ – Chris Eagle Feb 13 '12 at 1:04
  • $\begingroup$ And there are only 27 elements of $\mathbb F_3(\beta)$ to check ... $\endgroup$ – hmakholm left over Monica Feb 13 '12 at 1:07
  • $\begingroup$ Hi Chris, Suppose $\phi: \mathbb{F}_3(\alpha)\rightarrow\mathbb{F}_3(\beta)$ is the isomorphism. Then $\phi (\alpha)$ must be an expression in terms of $\beta$. Am I thinking about this the right way?!!? $\endgroup$ – bonyankan Feb 13 '12 at 1:11
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    $\begingroup$ I removed the "extraordinary isomorphisms" tag because for one, I feel like this is a rather ordinary isomorphism, and for two, I've never even heard that term and don't feel like it merits its own tag at this point. If someone feels this was hasty, please let me know. $\endgroup$ – NKS Feb 13 '12 at 1:22
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One way to do it is just to slug through it:

An isomorphism $\mathbb{F}_3(\alpha)\to\mathbb{F}_3(\beta)$ is completely determined by the image of $\alpha$; and, as you note in comments, the image of $\alpha$ must be of the form $a\beta^2 + b\beta + c$, with $a,b,c\in\mathbb{F}_3$. Moreover, the image of $\alpha$ must satisfy $\alpha^3 = \alpha - 1$; that is, you want to find $a,b,c$ such that $$(a\beta^2 + b\beta + c)^3 = a\beta^2 + b\beta + c-1.$$ So you can just expand the left hand side, using $\beta^3 = \beta^2-1$, and figure out the coefficients. We have: $$\begin{align*} \beta^3 &= \beta^2 - 1\\ \beta^4 &= \beta^3-\beta\\ &= \beta^2 - \beta - 1\\ \beta^6 &= (\beta^3)^2 = \beta^4 - 2\beta^2 + 1\\ &= \beta^2 - \beta - 1 -2\beta^2 + 1\\ &= -\beta^2 - \beta. \end{align*}$$ And so, since we are in characteristic $3$, $$\begin{align*} (a\beta^2 + b\beta + c)^3 &= a^3\beta^6 + b^3\beta^3 + c^3\\ &= a^3(-\beta^2-\beta) + b^3(\beta^2-1) + c^3\\ &= (b^3-a^3)\beta^2 + (-a^3)\beta + c^3-b^3\\ &= a\beta^2 + b\beta + c - 1. \end{align*}$$ So we need to solve the equations $$\begin{align*} b^3-a^3 &= a\\ -a^3 &= b\\ c^3 -b^3&= c-1. \end{align*}$$ Since $a,b,c\in\mathbb{F}_3$, where $x^3=x$ for all $x$, we get $$\begin{align*} b-a &= a\\ -a&=b\\ c-b&= c-1 \end{align*}$$ The first two equations both give $b=-a$; the last equation gives $b=1$. So $a=-1$, $b=1$, and $c$ is free (this gives the three roots of the polynomial). That is, $f(\alpha)$ can be any of $-\beta^2+\beta$, $-\beta^2+\beta+1$, $-\beta^2+\beta-1$. You can verify they all work.

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From $\beta^3 - \beta^2 + 1 = 0$, we get $1 - \beta^{-1} + \beta^{-3} = 0$, and since $\alpha^3 - \alpha + 1 = 0$, we can map $\alpha$ to $\beta^{-1}$ to get the desired isomorphism. Of course, from $\beta^3 - \beta^2 + 1 = 0$ we get $\beta^2 - \beta + \beta^{-1} = 0$, that is, $\beta^{-1} = -\beta^2 + \beta$. In other words, we can take $f(\alpha) = -\beta^2 + \beta$ exactly as Arturo Magidin found.

Edit: Added note We could instead have chosen to map $\alpha$ to the conjugates $\beta^{-3}$ or $\beta^{-3^2}$ of $\beta^{-1}$ where we have that $$\beta^{-3} = \beta^{-1} - 1 = -\beta^2 + \beta - 1,$$ and $$\beta^{-3^2} = (\beta^{-1} - 1)^3 = \beta^{-3}-1 = -\beta^2 + \beta + 1$$which are the other two possible images of $\alpha$ given in Arturo's answer. As in implied in Jyrki Lahtonen's comment, a simpler derivation is possible in this case because $\alpha$ and $\beta$ are roots of reciprocal polynomials.

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    $\begingroup$ +1: Stare at the minimal polynomials. Observe that they are each others reciprocal polynomials.... $\endgroup$ – Jyrki Lahtonen Feb 13 '12 at 7:09

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