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  "body": "\n## Definition\n\nLet $n \\ge 1$ be an integer, and let $K$ be a [[field|04C5ZBG1WT]] whose characteristic does not divide $n$.\nAssume that $K$ contains the group $\\mu_n$ of $n$ distinct $n$-th [[roots of unity|0YV2CKP2AV]].\nFix an algebraic closure $\\overline{K}$ of $K$.\nFor a subgroup $B$ of $K^\\times$ that contains $K^{\\times n} = \\set{x^n : x \\in K^\\times}$, let $K(\\sqrt[n]{B})$ be the subfield of $\\overline{K}$ generated over $K$ by all $\\beta \\in \\overline{K}$ with $\\beta^n \\in B$.\nA *Kummer extension* of exponent $n$ of $K$ is an extension of the form $K(\\sqrt[n]{B})$.\nFor such an extension $L = K(\\sqrt[n]{B})$, the *Kummer pairing* is the map\n$$\\Gal(L/K) \\times B/K^{\\times n} \\to \\mu_n, \\qquad \\langle \\sigma, b \\rangle = \\frac{\\sigma(\\beta)}{\\beta},$$\nwhere $\\beta \\in L$ is any element with $\\beta^n = b$.\nKummer theory is the study of these extensions through this pairing.\n\n## Remarks\n\nThe pairing is well defined: two choices of $\\beta$ differ by a factor in $\\mu_n \\subseteq K$, which $\\sigma$ fixes.\nThe main theorem of Kummer theory: $L/K$ is a [[Galois extension|0QMZAQ6QQ4]] with [[abelian|082GHNPJ53]] Galois group of exponent dividing $n$, and if $B/K^{\\times n}$ is finite the pairing is perfect, so $\\Gal(L/K) \\cong \\Hom(B/K^{\\times n}, \\mu_n)$.\nThe map $B \\mapsto K(\\sqrt[n]{B})$ is a bijection from the subgroups $B$ with $K^{\\times n} \\subseteq B \\subseteq K^\\times$ and $B/K^{\\times n}$ finite onto the finite abelian extensions of $K$ in $\\overline{K}$ of exponent dividing $n$.\nIn particular, for $a \\in K^\\times$, the extension $K(\\sqrt[n]{a})/K$ is [[cyclic|0KVGH0J5B3]] of degree equal to the order of $a$ in $K^\\times / K^{\\times n}$.\n\nExample: $K = \\Q$, $n = 2$, $\\mu_2 = \\set{\\pm 1}$, and $B$ generated by $\\Q^{\\times 2}$, $2$ and $3$.\nThen $L = \\Q(\\sqrt{2}, \\sqrt{3})$, and $B/\\Q^{\\times 2}$ has order $4$, so $\\Gal(L/\\Q) \\cong (\\Z/2\\Z)^2$, as expected.\n\nNon-example: $K = \\Q$ and $n = 3$.\nThe field $\\Q$ does not contain the three cube roots of unity, so the definition does not apply, and indeed $\\Q(\\sqrt[3]{2})/\\Q$ is not even Galois.\n",
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