Finding the ideals of an algebra

  • Thread starter Thread starter DeldotB
  • Start date Start date
  • Tags Tags
    algebra rings
Click For Summary
To find the ideals in the algebra \mathbb{Z}_5[x]/I, where I is generated by x^2 + 4, it is essential to clarify that the algebra A refers to \mathbb{Z}_5[x]/I. An ideal S in this algebra must satisfy the condition that the product of any element in S with any element in A remains in S. The polynomials in I are specifically of the form (x^2 + 4)Z_3[x], where Z_3[x] consists of polynomials of degree 3 or less with integer coefficients. Understanding these definitions is crucial for identifying the ideals within the algebra.
DeldotB
Messages
117
Reaction score
8
Say I have the algebra \mathbb{Z}_5[x]/I where is the I is the principle ideal generated by x^2+4. How do I find the ideals in A? I can't seem to find an explanation that is clear anywhere. Thanks!
 
Last edited:
Physics news on Phys.org
anyone?
 
I am a bit confused by your "How do I find the ideals in A?" when there is no mention of A! Did you mean A= Z_5 [x]/I?

A subset, S, of an algebra, A, is an "ideal" if and only if the product of a member of s with any member of A is again in S. Z_5[x] is the set of all polynomials of degree 5 or less with integer coefficients. I is the set of all such polynomials of the form (x^2+ 4)Z_3[x] where Z_3[x] is the set of all polynomials of degree 3 or less with integer coefficients.
 
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 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
847
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 5 ·
Replies
5
Views
917
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K