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

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Discrepancies between the rational root test and modulo p methods arise due to the irreducibility conditions outlined in Theorem 6.2.13, which indicates that a reducible polynomial in $\Bbb{Z}_p[x]$ does not imply reducibility in $\Bbb{Q}[x]$. The rational root test correctly demonstrated that the polynomial is irreducible in $\Bbb{Q}[x]$, while the modulo p method provided a one-way test that cannot confirm reducibility. Clarifications were sought regarding the interpretation of reducibility in $\Bbb{Q}[x]$ and the conditions for expressing polynomials as products of irreducible factors. The discussion emphasized that irreducibility prevents any factorization into polynomials of lower degree. Understanding these distinctions is crucial for correctly applying these methods in polynomial analysis.
Joe20
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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.
 

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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.
 

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