MHB Elements of a Ring: R Has 64 Elements

  • Thread starter Thread starter AkilMAI
  • Start date Start date
  • Tags Tags
    Elements Ring
Click For Summary
The discussion focuses on a ring homomorphism f: R->S, where the kernel of f contains 4 elements and the image has 16 elements. Using the relationship between the image, kernel, and the original ring, it is established that the number of elements in R is calculated as |R| = |ker f| * |Im(f)|. This leads to the conclusion that |R| = 4 * 16, resulting in R having 64 elements. The logic and calculations presented are affirmed as correct.
AkilMAI
Messages
73
Reaction score
0
f:R->S is a homomorphism of rings,such that kernel of f has 4 elements and the image of f has 16.How many elements has R?
16=|Im ( f )|=|R/ker f|=|R|/|ker f|=|R|/4=>|R|=4*16=64
[FONT=MathJax_Math]
 
Physics news on Phys.org
James said:
f:R->S is a homomorphism of rings,such that kernel of f has 4 elements and the image of f has 16.How many elements has R?
16=|Im ( f )|=|R/ker f|=|R|/|ker f|=|R|/4=>|R|=4*16=64

Yes, your logic and answer are both correct.
 
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 5 ·
Replies
5
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 41 ·
2
Replies
41
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
907
  • · Replies 3 ·
Replies
3
Views
837
  • · Replies 21 ·
Replies
21
Views
1K
Replies
1
Views
1K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K