If in standard algebra
every finite division ring is a field
from a superalgebra point of view what is the correspondent formulation to say that every "super-finite-division-ring" is a superfield ?
In Wedderburn's little theorem say that
every finite domain is a field is essentially equivalent to saying that the Brauer group of a finite field is trivial.
So if also supergroups exists what is generalization from a superalgebra point of view of the phrase "Brauer group of a finite field is trivial" ?
I change the height of the interval of the point of view (from Upper half-plane H instead of $R^2$-Euclidean plane) so I don't start from lower position (from ground to underground = R-plane) to build same theorem
ring
division-ring
field
group
but from higher position (from sky to ground "storey plan" not from sky to underground interval) = Upper half-plane $H$.
superring
(super)division-ring
superfield
supergroup
to find an equivalent theorem
NOTE:
R-plane is classical two-dimensional Euclidean space denoted $R^2$
A "storey plan" is any level part of a building but in this case 'bulding' is our structure
Sorry if I use a familiar image to explain a mathematical concept.