# What does the German word "Zerlegungsautomorphismus" translate to?

I would like to know if any of our German friends can translate that word for me.

Zerlegung is factorisation, isn't it? So what is factorisation automorphism?

This is taken from Deuring's paper “Die Zetafunktion einer algebraischen Kurve vom Geschlechte Eins (vierte Mitteilung)”. This is not the run of the mill question, hopefully it is still a valid question.

This is possibly the first time this term appears, although this is part 4 of a series of paper, so I'm not sure if he has used/defined it in the earlier parts.

Wir betrachten den Fall, daß p in $$k_1$$ prim bleibt, $$\mathbf{p}=\mathbf{P},\quad\mathbf{p}^{\varphi}=\mathbf{P}.$$ $$\varphi$$ ist dann Zerlegungsautomorphismus von $$\mathbf{P}$$ über $$k$$, also auch von $$\mathbf{P}_{\Sigma}$$ über dem Körper $$P$$ der rationalen Zahlen, $$\mathbf{p}=\mathbf{P}_{\Sigma}\quad\text{oder}\quad\mathbf{p}^{\varphi}=\mathbf{P}.$$

Danke sehr!

• Zerlegung could as well mean partitoining ... Jul 12, 2013 at 13:51
• But what would a partitioning automorphism mean though? Jul 12, 2013 at 13:51
• I tried googling "Zerlegungsautomorphismus", which looks (to me) more like a German word than "Zerlegungautomorphism", but I got nothing useful. My best guess is that it means an automorphism of some decomposition, i.e., a bijection that sends some decomposition (of something) to itself, but this is only a guess. Jul 12, 2013 at 14:45
• If no one answers, maybe provide a quote of a few sentences before the first time it is used in the paper. Jul 12, 2013 at 15:45
• Maybe "decomposition" instead? There is a decomposition group and a decomposition field. If the group is cyclic, then it'd make sense to call its generator a decomposition automorphism. Jul 12, 2013 at 16:33

Even in English texts the symbol $Z$ (resp. $T$) often stands for the decomposition group (resp. inertia group). I guess this is a tribute to the contributions of German number theorists. We can then identify $Z/T$ as the Galois group of the related extension of residue class fields. In the listed cases the residue class fields are finite, so $Z/T$ is then necessarily cyclic.
• The Galois group of the residue field extension is cyclic, and lifts to a cyclic-group-mod-$T$, yes. Frobenius. Jul 12, 2013 at 16:46