/sci/ - Science & Math

>>2558059
can you rephrase your question so that it makes more sense? also, you don't get a field when looking at integers with modulo 4 arithmetic since 4 is not a prime number...

Guys I don't really have an idea what the fuck are you talking about. I'll just go watch Pulp Fiction now.

>>2558078
well I suppose my question is:
How does addition arithmetic work in a finite field of say F4? in other notation that would be GF(2^2)
How does multiplication arithmetic work in GF(2^2) ?

I know the addition and multiplication tables are the same for GF(Prime) and Modulo Prime...

le bump

One of the axioms for it being in a field is that it must be closed under addition and multiplication; i.e adding two elements always gives another. When you do such binary operations you remain in the field. Then the multiplication tables are build from there using the idea of a modular arithmetic

>>2558341

*for it being A field

My bad.

>>2558110
um, it works how you define it to work, by a table, or some set of rules

so long as your defn doesn't conradict the field axioms of course

>>2558341
sure, i get that on the surface

i'm talking about the actual calculations for GF(2^2) of 1 + 1 = 0 and 2 + 2 = 0 and 3 + 3 = 0

>>2558362
they are that way by definition

>>2558059
>what I dont get is how 1 + 1 in a finite field F4 = 0 or 2 + 2 = 0.

That's because the characteristic is 2.

>>2558367
>>2558373

so because
1+1 = an even number
and 2+2 = an even number
and 3+3 = an even number
they equal 0 because 2 is the characteristic ?

Which GF(4) are you working with?

>>2558410
well, yes.

or you could say the characteristic is 2 because 1 + 1 = 0, 2 + 2 = 0, 3 + 3 = 0

i think you need to get over the what it all means business and start thinking in the abstract

>>2558410
Yes.

>>2558412
They are all isomorphic, so they're considered all the same.

>>2558419
>>2558429
ok, i'm starting to see that.

well, how about in GF(2^2), 3 + 1 = 2? what is the logic behind that?

>>2558512
because if 3 + 1 = 0 then 3 + 1 + 1 = 1, and because characteristic is 2, 1 + 1 = 0 so 3 + 0 = 1, ie 3 = 1. impossible.

and if 3 + 1 = 1 then 3 would be the additive identity 0

if 3 + 1 = 3 then 1 would be additive identity 0

these results are forced on us by the fact that it is a field

>>2558429
Yes I know they are all isomorphic, but I wanted to give a concrete example as to why 1+1 = 0 in fields of characteristic 2.

>>2558571
plus, while it makes sense to talk about 0 and 1 in GF(4), it makes no sense to talk about "2" or "3" without giving the appropriate irreducible polynomial you used to make your field extension.

>>2558571
> concrete example

i suggest you either give that up or give up abstract algebra

it is called abstract for a reason

>>2558575
i meant concrete as in thinking of elements of GF(4) as elements in GF[\alpha], and showing explicity how polynomial arithmetic imply what OP is stating.

>>2558571
>why 1+1 = 0 in fields of characteristic 2.

lol, you stupid retard

why is the number 7 the number 7? i just don't get this maths business

>>2558574
finite fields don't have to come from polynomials.

next you'll be asking why i and -i are both the sqrt of -1 though

>>2558581
err, yes i get the definition of characteristic. But this can be shown directly from polynomial arithmetic and the arithmetic of GF(2), which the OP knows and understands.

>>2558584
yes for sure you could talk about GF(4) in a basis-less setting, but you can't talk about "2" or "3"....f

>>2558589
and before you try to get me, yes i know only one GF(4).