Let ${\displaystyle L/K}$  be a finite Galois extension of global fields with Galois group ${\displaystyle G}$ . Then the discriminant equals

${\displaystyle {\mathfrak {d}}_{L/K}=\prod _{\chi \in \mathrm {Irr} (G)}{\mathfrak {f}}(\chi )^{\chi (1)},}$ 

where ${\displaystyle {\mathfrak {f}}(\chi )}$  equals the global Artin conductor of ${\displaystyle \chi }$ .^[1]

Let ${\displaystyle L=\mathbf {Q} (\zeta _{p^{n}})/\mathbf {Q} }$  be a cyclotomic extension of the rationals. The Galois group ${\displaystyle G}$  equals ${\displaystyle (\mathbf {Z} /p^{n})^{\times }}$ . Because ${\displaystyle (p)}$  is the only finite prime ramified, the global Artin conductor ${\displaystyle {\mathfrak {f}}(\chi )}$  equals the local one ${\displaystyle {\mathfrak {f}}_{(p)}(\chi )}$ . Because ${\displaystyle G}$  is abelian, every non-trivial irreducible character ${\displaystyle \chi }$  is of degree ${\displaystyle 1=\chi (1)}$ . Then, the local Artin conductor of ${\displaystyle \chi }$  equals the conductor of the ${\displaystyle {\mathfrak {p}}}$ -adic completion of ${\displaystyle L^{\chi }=L^{\mathrm {ker} (\chi )}/\mathbf {Q} }$ , i.e. ${\displaystyle (p)^{n_{p}}}$ , where ${\displaystyle n_{p}}$  is the smallest natural number such that ${\displaystyle U_{\mathbf {Q} _{p}}^{(n_{p})}\subseteq N_{L_{\mathfrak {p}}^{\chi }/\mathbf {Q} _{p}}(U_{L_{\mathfrak {p}}^{\chi }})}$ . If ${\displaystyle p>2}$ , the Galois group ${\displaystyle G(L_{\mathfrak {p}}/\mathbf {Q} _{p})=G(L/\mathbf {Q} _{p})=(\mathbf {Z} /p^{n})^{\times }}$  is cyclic of order ${\displaystyle \varphi (p^{n})}$ , and by local class field theory and using that ${\displaystyle U_{\mathbf {Q} _{p}}/U_{\mathbf {Q} _{p}}^{(k)}=(\mathbf {Z} /p^{k})^{\times }}$  one sees easily that if ${\displaystyle \chi }$  factors through a primitive character of ${\displaystyle (\mathbf {Z} /p^{i})^{\times }}$ , then ${\displaystyle {\mathfrak {f}}_{(p)}(\chi )=p^{i}}$  whence as there are ${\displaystyle \varphi (p^{i})-\varphi (p^{i-1})}$  primitive characters of ${\displaystyle (\mathbf {Z} /p^{i})^{\times }}$  we obtain from the formula ${\displaystyle {\mathfrak {d}}_{L/\mathbf {Q} }=(p^{\varphi (p^{n})(n-1/(p-1))})}$ , the exponent is

${\displaystyle \sum _{i=0}^{n}(\varphi (p^{i})-\varphi (p^{i-1}))i=n\varphi (p^{n})-1-(p-1)\sum _{i=0}^{n-2}p^{i}=n\varphi (p^{n})-p^{n-1}.}$ 

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