MHB Proving Two Rings Are Isomorphic

  • Thread starter Thread starter cs0978
  • Start date Start date
  • Tags Tags
    Rings
cs0978
Messages
1
Reaction score
0
I seem to be having a difficult time trying to figure out how to do this problem. It's from a non-graded homework assignment. I was able to get every other problem except for this one.Let R = Z, together with the two operations:
a + b := a + b + 3 and ab := ab + 3a + 3b + 6

Let S = Z, together with the two operations:
a + b := a + b - 2 and ab := -ab + 2a + 2b - 2

Assume ordinary addition and multiplication in Z and that R and S are rings.

Prove that R is isomorphic to S.
 
Physics news on Phys.org
cs0978 said:
I seem to be having a difficult time trying to figure out how to do this problem. It's from a non-graded homework assignment. I was able to get every other problem except for this one.Let R = Z, together with the two operations:
a + b := a + b + 3 and ab := ab + 3a + 3b + 6

Let S = Z, together with the two operations:
a + b := a + b - 2 and ab := -ab + 2a + 2b - 2

Assume ordinary addition and multiplication in Z and that R and S are rings.

Prove that R is isomorphic to S.
As a start, find what the zero and unit elements are in $R$ and $S$. An isomorphism must take the zero element of $R$ to the zero element of $S$, and the same for the unit elements.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
Replies
6
Views
1K
  • · Replies 5 ·
Replies
5
Views
793
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
937
  • · Replies 2 ·
Replies
2
Views
3K
Replies
1
Views
1K
Replies
6
Views
2K