Search the corpus

    Possible Galois groups of an abelian extension of degree $540$, and quadratic intermediate extensions $L/E$

    P-4ZDQQ

    Area
    Algebra
    Topics
    Abelian Groups, Field Extensions, Galois Theory
    Status
    draft
    • If \(L/k\) is an abelian Galois extension of degree \(540 = 2^2 \times 3^3\times 5\), what are the possible Galois groups \(\Gal(L/k)\)?

      • Are there any intermediate fields \(E\) for which \(L/E\) is a quadratic extension?
    Solution

    Let \[ G=\operatorname{Gal}(L/k). \] Since \(L/k\) is abelian Galois and \([L:k]=540\), \(G\) is a finite abelian group of order \[ 540=2^2 3^3 5. \]

    1. The possible primary components are\[ G_{(2)}\cong C_4\quad\text{or}\quad C_2\oplus C_2, \]\[ G_{(3)}\cong C_{27},\quad C_9\oplus C_3,\quad\text{or}\quad C_3\oplus C_3\oplus C_3, \]and\[ G_{(5)}\cong C_5. \]

    These are exactly the abelian groups corresponding to the partitions of the exponents \(2\), \(3\), and \(1\) in the primary decomposition theorem.

    2. Hence there are exactly six possible Galois groups.

    The \(2\)-primary and \(3\)-primary choices are independent, while the \(5\)-primary part is forced. Thus there are \[ 2\cdot3\cdot1=6 \] isomorphism classes. In invariant-factor form they are \[ C_{540}, \qquad C_3\oplus C_{180}, \qquad C_3\oplus C_3\oplus C_{60}, \] \[ C_2\oplus C_{270}, \qquad C_6\oplus C_{90}, \qquad C_3\oplus C_6\oplus C_{30}. \]

    3. An intermediate field \(E\) satisfies \([L:E]=2\) exactly when the corresponding subgroup \(\operatorname{Gal}(L/E)\le G\) has order \(2\).

    By the fundamental theorem of Galois theory, \[ [L:E]=|\operatorname{Gal}(L/E)|. \] Thus quadratic extensions \(L/E\) correspond precisely to order-\(2\) subgroups of \(G\).

    4. Such intermediate fields always exist.

    Since \(2\mid |G|\), Cauchy’s theorem gives an element of order \(2\), hence a subgroup of order \(2\). By 3 this subgroup corresponds to an intermediate field \(E\) with \([L:E]=2\).

    5. The number of such fields is determined by the \(2\)-primary part.

    If \(G_{(2)}\cong C_4\), there is a unique subgroup of order \(2\), hence exactly one such \(E\). If \(G_{(2)}\cong C_2\oplus C_2\), its three nonzero elements generate three distinct order-\(2\) subgroups, hence there are exactly three such fields. In every case \[ [E:k]=[L:k]/2=270. \]