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Let $f= x^5 − x − 1$ and $L$ be the splitting field of $f(x)$ over $\Bbb Q$. Suppose $B$ is an integral closure of $\Bbb Z$ in $L$ and $P$ is a maximal ideal of $B$ such that $P \cap \Bbb Z = 2\Bbb Z$. I am trying to compute the Galois group of $f$, i.e., the group $Gal(L/\Bbb Q)$, using the following theorem:

Theorem. Let $A$ be an entire ring, integraly closed in its quotient field $K$. Let $L$ be a finite Galois extension of $K$. Let $K(\alpha)$, where $\alpha$ is integral over $A$, and let $f(x)=x^n+a_{n-1}x^{n-1}+...+a_0$ be the irreducible polynomial of $\alpha$ over $K$ with $a_i\in A$. Let $\mathfrak{p}$ be a maximal ideal in $A$, let $\mathfrak{B}$ be a prime ideal of the integral closure $B$ of $A$ in $L$, $\mathfrak{B}$ lying over $\mathfrak{p}$. Let $\bar{f}$ be the reduced polynomial with coefficients in $A/\mathfrak{p}$. Let $G_{\mathfrak{B}}$ be the decomposition group. If $\bar{f}$ has no multiple roots, then the map $\sigma \mapsto \bar{\sigma}$ is an isomorphism of $G_{\mathfrak{B}}$ on the Galois group of $\bar{f}$ over $A/\mathfrak{p}$.

I see that the theorem can be applied with $A=\Bbb Z, K=\Bbb Q, L=L, \mathfrak{p}=2\Bbb Z$, and $\mathfrak{B}=P$, but I can't see how to compute $G_\mathfrak{B}$ or the Galois group of $\bar{f}$ over $A/\mathfrak{p}=\Bbb Z/2\Bbb Z$. I see that $\bar{f}$ is a product of two irreducible polynomials in $\Bbb Z/2\Bbb Z[x]$, of degrees $2$ and $3$. Also how can I deduce the Galois group $Gal(L/\Bbb Q)$ by the theorem?

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The fact that $f$ decomposes that way when reduced modulo 2 tells you that the Galois group $G$ of $f$, which is a subgroup of $S_5$ since $\deg f=5$, contains an element $\sigma$ that decomposes as a product of a 2-cycle and a 3-cycle. Hence, $G$ contains a transposition, namely $\sigma^3$, and a 3-cycle, namely $\sigma^2$. I'm not sure whether this is already enough to conclude that $G=S_5$, but $f$ is irreducible mod 3, and therefore $G$ also contain a 5-cycle. Since $S_p$ is generated by a $p$-cycle and a transposition for every prime $p$, you get that $G=S_5$.

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