Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This appears in Dickson's "Linear Groups with an exposition of the Galois field theory", page 50, chapter 4.

enter image description here enter image description here

My question is: how would the above "translate" in modern terms? In particular, do we have a name for this multiplier Galois field?

share|improve this question
"Scalar" perhaps? –  anon Apr 1 '12 at 9:09
I haven't included what Dickson writes next, but he goes on to show that these "multipliers" actually form a subfield of $F_{p^n}$ which he calls the multiplier Galois field of the additive group $[\lambda_1,...,\lambda_m]$. Now that I think of it, it's actually this thing I wanted to ask about, whether it has a special name that is. –  the_fox Apr 1 '12 at 9:21
Just edited pictures and question. –  the_fox Apr 1 '12 at 9:28
What does Dickson do with this concept later? His term "marks" is an old way of saying "elements". He first defines the additive group $V = {\mathbf F}_p^m$, and to make sense of multiplying elements of $V$ by elements of ${\mathbf F}_{p^n}$ in a natural way it seems like one should view $V$ inside $W = {\mathbf F}_{p^n}^m$. But then the only $\mu$ that fit his condition would be elements of ${\mathbf F}_p$, which can't be what he intends. Hmm. What he is doing seems similar to the definition of the multiplier of a lattice in algebraic number theory: if $K$ is a number field (contd..) –  KCd Apr 1 '12 at 12:09
of degree $n$ over ${\mathbf Q}$ and $M$ is a full lattice in $K$ (meaning a ${\mathbf Z}$-module of rank $n$ in $K$) then the multiplier ring of $M$ is ${\mathcal O}_M = \{\alpha \in K : \alpha{M} \subset M\}$. This is an order in $K$ (not necessarily the maximal order = alg. integers of $K$) and it's a useful concept. Back in Dickson's setting, his condition ${\mu}V \subset V$ looks like $\alpha{M} \subset M$, so you can a similarity, but I already indicated that it seems like Dickson's thing is just leading to ${\mathbf F}_p$, which has to be a misunderstanding on my part. –  KCd Apr 1 '12 at 12:13

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.