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14555024 No.14555024 [Reply] [Original]

Brainlet here, R cross R is an R-algebra, being a cross product of (trivial) R algebras. It's clearly two dimensional. but this:

https://math.stackexchange.com/questions/2217906/what-are-the-three-non-isomorphic-2-dimensional-algebras-over-mathbbr

shows that all 2d R algebras are either the complex numbers, the set A = {x + jy} where j^2 = 1 or B = {x + ey} where e^2 = 0. Which one of these is RxR then?

>> No.14555027

>>14555024
This is the fourth time I see you make this thread. You get the same answer every time, that it's A. Why do you keep posting this?

>> No.14555034

>>14555027
schizo

>> No.14555036

>>14555034
Who? Me or OP?

>> No.14555051

>>14555024
why the fuck do you keep posting this

>> No.14555139

Regard the elements of RxR as ordered pairs. Then just have multiplication be given coordinatewise.
We now have (1,0)^2=(1,0) and (0,1)^2=(0,1). Now you can just verify that the map sending (1,0) to 1 and (0,1) to j is an isomorphism

>> No.14555147

>>14555139
That's wrong though. (0,1)^2 != (1,0)

>> No.14555251

What year would this be covered for a math degree (at the earliest)? I'm curious.

>> No.14555254

>>14555251
First year. You can literally do this in linear algebra.