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Let $L/K$ be an extension of number fields and $v$ be a prime (an equivalence class of valuations) of $K$ and $d_L$ the absolute discriminant of $L$.

I know that a rational prime $p$ in $\mathbb Q$ ramifies in $L$ iff $p\mid d_L$.

Is it true that an archimedean prime $v$ ramifies in $L$ if and only if $v\mid d_L$? How can I prove this?

Also then what is the significance of a product of some collection of algebraic objects taken over such $v$'s, for example, something like $\prod_{v\mid d_L}A_{v,w}$ with $A_{v,w}$ complex numbers where for each $v$ a prime $w$ of $L$ is chosen above $v$?

Many thanks for your help.

eddie
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    A place $v$ of $K$ ramifies in $L$ iff $v$ divides the relative discriminant $\mathfrak{d}_{L/K}$ of the extension $L/K$. When $K = \mathbb{Q}$, this coincides with the absolute discriminant (or rather it is the ideal in $\mathbb{Z}$ generated by the absolute discriminant). – Peter Humphries Apr 28 '16 at 22:28
  • So if $v\mid d_L$ then does this tell us anything about $v$ in the case $K\neq \mathbb Q$? Also what happens if $v$ is archimedean? – eddie Apr 28 '16 at 23:15
  • You're asking the wrong question. $d_L$ is the cardinality of $\mathfrak{d}{L/\mathbb{Q}}$. We have that $\mathfrak{d}{L/\mathbb{Q}} = N_{K/\mathbb{Q}}(\mathfrak{d}{L/K}) \mathfrak{d}{K/\mathbb{Q}}^{[L \colon K]}$. (This is true with $L,K,\mathbb{Q}$ replaced by any tower of local fields too.) – Peter Humphries Apr 28 '16 at 23:22
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    In particular, suppose that a place $p$ of $\mathbb{Q}$ ramifies in $K$. Let $\mathfrak{p}$ lie over $p$ in $K$. Then $\mathfrak{p}$ need not necessarily ramify in $L$, yet $p \mid d_L$. – Peter Humphries Apr 28 '16 at 23:24
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    Dear @eddie, Note that an Archimedean prime is an equivalence class of absolute values. It makes no sense to talk about an Archimedean prime dividing an integer. See this post for ramification of Archimedean primes: http://math.stackexchange.com/questions/22620/what-does-it-mean-for-a-prime-at-infinity-to-ramify. – Keenan Kidwell Apr 28 '16 at 23:47

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