Recent content by Stephen88

Re: Ring and cosest So I need to apply the division algorithm on (AB)+I...ok

I'm trying to list the cosets of the following ring and describe the relations that hold between these cosets. R=Z_4[x]/((x^2+1)*Z_4[x]) I'm using the division algorithm since x^2+1 is monic in the ring Z_4[x].Now for every f that belongs to Z_4[x] by the division algorithm...

I'm trying to see if the map f:C->Z,f(a+bi)=a is a homomorphism of rings. Let x=a+bi and y=c+di...then f(x+y)=a+c=f(x)+f(y) but f(xy)=f((ac-db)+(ad+bc)i)=ac-db/= f(x)f(y)...so the map is not a homomorphism of rings. Is this correct?

Yes after adding the additive inverse on both sides I get 0 on the left and 'a' on the right (since a*1=a*0=a. )

Homework Statement 1_R=identity in the ring R. /=...not equal Having some issues with this any help will be great: Let R be a ring with identity, such that x^2 = 1_R for all 0_R /= x ,where x belongs to R. How many elements are in R? Homework Equations The Attempt at a Solution...

Well...Since a*1=a by the axiom and a*0=0 because I'm multiplying with the additive identity or 0 then a=0...I guess..I'm walking in circles

Ok ,by Axiom number 3.. a+0=a which can be written as a+0=a=a*1 which is true because of Axiom number 8 ..but 0=1 therefore a*1=a*0>a=0

because 0 is the additive identity,the multiplicative zero so by definiton a*0=0...and 1 is the multiplicative identity so by definition a*1=a...

Very true.Ok let me try again: a*0 is always 0 and a*1 is always a ...but in this context 0 = 1 therefore 0=1>a*0=a*1=a>0=a

Yes normally 0 is the additive identity and 1 the multp. one...but here they are both the same. So if 1=0->1*a=0*a=a=>a=1=0? If not I'm confused

a product of a and the multiplicative identity?

Uh sorry about that...I;m used to integers and real numbers...hmm..given the context both represent the identity of the ring so a*0=a*1=a

yes a = 0 is something I need to prove that was just I way that I thought it would be possible. For instance a*0=a*(0+0)=a*0+a*0=a*1+a*1=2a=>a*0=2a= >a=0..otherwise 2=0 which is a contradiction.Is this correct?

0=0+0=>a*0=a*(0+0)=a*0+a*0=a*1_r+a*1_r=2a=>a*0=2a=>a=0...is this correct?

Homework Statement Let R be a ring in which 1_R = 0_R .Show that R has only one element.Homework Equations The Attempt at a Solution I'm trying to show that a*0_r=a*1_r implies a*0_r=0_r. if 0=0+0=>a*0=a*(0+0)=a*0+a*0=a*1_r+a*1_r=2a=>a*0=2a= >a=0...is this correct? If not Is there something I...