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Compute the Galois group for the root field of the polynomial $x^3$+$2x$+2 over $Z_3$

Choose a root $\alpha$ in the root field of the polynomial.

We found three roots in this root field, $\alpha$, $\alpha$+1, $\alpha$+2.

What are the automorphisms then? Is the Galois group just these three? Or is it just the first 2 since you can generate the third one with the second one?

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The elements of the Galois group are not elements of the field. You may take one element of the group to be the one that sends $\alpha$ to $\alpha+1$. This is often designated $\alpha\mapsto\alpha+1$. When you do it twice, the $1$ remains fixed, as it must, and the $\alpha$ of the image goes to $\alpha+1$, so the upshot is $\alpha\mapsto\alpha+2$. Do it once more and $\alpha\mapsto\alpha$, the identity. So you have $3$ things in the Galois group, as you must, for a normal cubic extension.

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  • $\begingroup$ Wow, posted within seconds of each other. Nice and concise! +1 :) $\endgroup$ – Alex Wertheim Jun 12 '14 at 21:57
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Let $f(x) = x^{3}+2x+2$ over $\mathbb{Z}_{3}$. $f$ is clearly irreducible over $\mathbb{Z}_{3}$, as it has no roots in $\mathbb{Z}_{3}$ - this can be checked easily. Let $\alpha$ be a root of $f(x)$, and consider the extension $\mathbb{Z}_{3}(\alpha)$. As you correctly noted, the other roots of $f(x)$ are given by $\alpha+1$ and $\alpha+2$. Hence, $\mathbb{Z}_{3}(\alpha)$ is the splitting field of $f(x)$.

From here, if you like, you can use the fact that $[\mathbb{Z}_{3}(\alpha):\mathbb{Z}_{3}] = 3$ (since $\deg(f(x)) = 3$) to conclude immediately that $\mathrm{Gal}(\mathbb{Z}_{3}(\alpha)/\mathbb{Z}_{3}) \cong \mathbb{Z}_{3}$.

If you want to more concretely answer the question, consider the possible automorphisms of your field extension. You have to map $\alpha$ to another root of $f$, so what are your options? There's clearly the identity map. Consider the map $\sigma$ which takes $\alpha$ to $\alpha+1$. Is this an automorphism? Consider now $\sigma^{2}$. Where does it take $\alpha$? $\alpha + 1$? See if you can show that $\langle \sigma \rangle = \{1, \sigma, \sigma^{2}\}$ is the complete automorphism group of this extension. Then, clearly, $\mathrm{Gal}(\mathbb{Z}_{3}(\alpha)/\mathbb{Z}_{3}) \cong \mathbb{Z}_{3}$. I'm happy to say more if you need clarification.

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