MHB What Causes Discrepancies Between Rational Root Test and Modulo p Methods?

Joe20
Messages
53
Reaction score
1
Hi all,

I have done the question in two methods. The first method is done by rational root test and the second method is by modulo p (theorem is as attached). It seems that my answers for both methods do not tally.

1. Where have I done wrong in the attached for the methods? Which is the correct presentation of answer for this question (i.e. rational test method or modulo p ?
2. How do I tell when to use rational root test method or modulo p method? When modulo p method not applicable?


Your advise is greatly appreciated. Thanks.
 

Attachments

  • Webp.net-resizeimage1.jpg
    Webp.net-resizeimage1.jpg
    84.1 KB · Views: 84
  • Webp.net-resizeimage2.jpg
    Webp.net-resizeimage2.jpg
    47.3 KB · Views: 103
  • Webp.net-resizeimage3.jpg
    Webp.net-resizeimage3.jpg
    70.8 KB · Views: 90
  • 3.png
    3.png
    8.1 KB · Views: 128
  • T1.png
    T1.png
    10.5 KB · Views: 108
Physics news on Phys.org
Your Theorem 6.2.13 says that if $\deg \overline{f}(x) = \deg f(x)$ and $\overline{f}(x)$ is irreducible in $\Bbb{Z}_p[x]$ then $f(x)$ is irreducible in $\Bbb{Q}[x]$. The theorem does not tell you anything at all about what might happen when $\overline{f}(x)$ is reducible in $\Bbb{Z}_p[x]$. So the fact that your $\overline{f}(x)$ is reducible in $\Bbb{Z}_3[x]$ does not imply that ${f}(x)$ is reducible in $\Bbb{Q}[x]$. That would amount to saying that the converse of Theorem 6.2.13 is true, which is certainly not the case.

The technique of looking at $\overline{f}(x)$ in $\Bbb{Z}_p[x]$ is a one-way test. You can use it to prove irreducibility of $f(x)$, but you cannot use it to prove reducibility.

Your argument using the rational root test is correct, and shows that $f(x)$ is irreducible in $\Bbb{Q}[x]$.
 
Thank you for the advice.

I have one more question. Can you kindly help me to check if the argument or reasoning stated in the last part of the rational root test method is correct? ( from this " f(x) has no linear factor, hence 2x^2+2x -3 ... till hence 2x^4+8x^3+5x^2-7x-3 = (2x^2+2x -3)(x^2+3x+1).

Thanks.
 
Alexis87 said:
Thank you for the advice.

I have one more question. Can you kindly help me to check if the argument or reasoning stated in the last part of the rational root test method is correct? ( from this " f(x) has no linear factor, hence 2x^2+2x -3 ... till hence 2x^4+8x^3+5x^2-7x-3 = (2x^2+2x -3)(x^2+3x+1).

Sorry, I forgot that your rational root investigation actually showed that $2x^4+8x^3+5x^2-7x-3$ is reducible in $\Bbb{Q}[x]$. You correctly showed that $2x^4+8x^3+5x^2-7x-3 = (2x^2+2x -3)(x^2+3x+1).$
 
I am somehow confused by the question with this statement " if it is reducible in Q[x], express it as a product of irreducible polynomials in Q[x]. "

1. Does the part "if it is reducible in Q[x]" refers to any factorization [i.e. like the case of 2x^4+8x^3+5x^2−7x−3=(2x^2+2x−3)(x^2+3x+1) ] without the need to solve for the x value to check if it rational or irrational? Then the next question is how does this apply to reducible in Q[x]?

2. The next part " express it as a product of irreducible polynomials in Q[x]. " refers to (2x^2+2x−3) and (x^2+3x+1) being degree 2 and no roots to each quadratic polynomial?

3. So it also implies that if the polynomial is irreducible, then I can't even get to any form of factorization, hence needless to say any product of polynomials. Am I correct to say that?

Thanks.
 
Opalg said:
Sorry, I forgot that your rational root investigation actually showed that $2x^4+8x^3+5x^2-7x-3$ is reducible in $\Bbb{Q}[x]$. You correctly showed that $2x^4+8x^3+5x^2-7x-3 = (2x^2+2x -3)(x^2+3x+1).$
I am somehow confused by the question with this statement " if it is reducible in Q[x], express it as a product of irreducible polynomials in Q[x]. "

1. Does the part "if it is reducible in Q[x]" refers to any factorization [i.e. like the case of 2x^4+8x^3+5x^2−7x−3=(2x^2+2x−3)(x^2+3x+1) ] without the need to solve for the x value to check if it rational or irrational? Then the next question is how does this apply to reducible in Q[x]?

2. The next part " express it as a product of irreducible polynomials in Q[x]. " refers to (2x^2+2x−3) and (x^2+3x+1) being degree 2 and no roots to each quadratic polynomial?

3. So it also implies that if the polynomial is irreducible, then I can't even get to any form of factorization, hence needless to say any product of polynomials. Am I correct to say that?

Thanks.
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top