MHB What Does Reducing Z[x] Modulo the Prime Ideal (p) in Polynomial Rings Mean?

Click For Summary
Reducing Z[x] modulo the prime ideal (p) results in the polynomial ring Z/pZ[x], where the coefficients are taken from the field Z/pZ. This process involves replacing the integer coefficients of the polynomials in Z[x] with their equivalence classes modulo p. The notation Z/pZ[x] is interpreted as the polynomial ring over the field Z/pZ, which can also be denoted as (Z/pZ)[x]. The discussion clarifies that the reduction refers to the coefficients of the polynomials rather than the polynomials themselves. Understanding this notation is crucial for grasping the properties of the resulting Principal Ideal Domain.
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading Dummit and Foote Section 9.2: Polynomial Rings Over Fields I

I am having some trouble understanding Example 3 on page 300 (see attached)

My problem is mainly with understanding the notation and terminology.

The start of Example 3 reads as follows.

"If p is a prime, the ring Z/pZ[x] obtained by reducing Z[x] modulo the prime ideal (p) is a Principal Ideal Domain, since the coefficients lie in the field Z/pZ ... ... "

To me the ring Z/pZ[x] would be formed by reducing Z modulo p to form three cosets, namely
png.latex
and then forming Z/pZ[x] by taking coeffiients from Z/pZ

I am really unsure what D&F mean by "reducing Z[x] modulo the prime ideal (p)" unless they mean reducing the coefficients of Z[x] to coefficients from Z/pZ.

I am also assuming that when D&F use the notation Z/pZ[x] they are meaning (Z/pZ)[x]Can someone clarify this for me?

Peter
 
Physics news on Phys.org
Peter said:
I am really unsure what D&F mean by "reducing Z[x] modulo the prime ideal (p)" unless they mean reducing the coefficients of Z[x] to coefficients from Z/pZ.

I am also assuming that when D&F use the notation Z/pZ[x] they are meaning (Z/pZ)[x]

I am pretty sure they mean just what you say, the polynomial ring $(\mathbb{Z}/p\mathbb{Z})[x]$ or with other notations, $\mathbb{Z}_p[x]$ or $\left(\mathbb{Z}/(p)\right)[x]$.
 
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
48
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 5 ·
Replies
5
Views
847
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
783
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
20K