Re: Subrings of F_p^n

On 24 Jan, 16:02, Harald Helfgott <harald.helfg...@xxxxxxxxx> wrote:
Hi,

Let F_p be the finite group with p elements.

What are the subrings of (F_p)^n = F_p x ... x F_p (n times)?

I'll assume you mean unital subrings.

For each n there is the standard prime subring of (F_p)^n
namely that consisting of all (a,a,...,a) for a in F_p.
Call this R_n.

If n = n_1 + ... + n_k, then
R_{n_1} x ... x R_{n_k} is a subring of (F_p)^n.
One can permute the coordinates in (F_p)^n and get a
subring corresponding to each set partition of {1,...,n}.
If we like, these are the functions {1,...,n} -> F_p
constant on each part of the partition.

These are the only unital subrings of (F_p)^n.
Given a in F_p there is a polynomial over Z taking a to 1 and
all other elements of F_p to zero. Annying this polynomial to
an element of a subring S of (F_p)^n gives an idempotent of
(F_p)^n lying in S. It follows that each element of S is
an F_p-linear combination of idempotents in S. Idempotents
correspond to subsets of {1,2,...,n} and the ones in S form
a Boolean algebra. Thus the primitive idempotents form
a set partition of {1,2,...,n}. This proves my assertion.

Victor Meldrew
"I don't believe it!"

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