MHB GCDs of Polynomials: Reading Rotman's Corollary 3.58

  • Thread starter Thread starter Math Amateur
  • Start date Start date
  • Tags Tags
    Polynomials
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading Joseph J.Rotman's book, A First Course in Abstract Algebra.

I am currently focused on Section 3.5 From Polynomials to Numbers

I need help with the statement and meaning of Corollary 3.58

The relevant section of Rotman's text reads as follows:https://www.physicsforums.com/attachments/4547In the above text (in the statement of the Corollary) we read the following:

" ... ... (ii) Every two polynomials $$f(x)$$ and $$g(x)$$ have a unique gcd. ... ... "
My question (which some may regard as pedantic :) ) is as follows:

Does Rotman actually mean ...

" (ii) Every two polynomials $$f(x)$$ and $$g(x)$$ have a unique monic gcd. ... ... "

Can someone please confirm that my interpretation is correct?

Peter
 
Physics news on Phys.org
Peter said:
I am reading Joseph J.Rotman's book, A First Course in Abstract Algebra.

I am currently focused on Section 3.5 From Polynomials to Numbers

I need help with the statement and meaning of Corollary 3.58

The relevant section of Rotman's text reads as follows:In the above text (in the statement of the Corollary) we read the following:

" ... ... (ii) Every two polynomials $$f(x)$$ and $$g(x)$$ have a unique gcd. ... ... "
My question (which some may regard as pedantic :) ) is as follows:

Does Rotman actually mean ...

" (ii) Every two polynomials $$f(x)$$ and $$g(x)$$ have a unique monic gcd. ... ... "

Can someone please confirm that my interpretation is correct?

Peter
I think gcd is defined to a monic by default. Please check the definition in the book. If it doesn't say monic specifically then yes, you can have many gcds.
 
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
1
Views
1K
  • · Replies 24 ·
Replies
24
Views
711
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
1K