Polynomials in Z6[x]: Find & Explain Deg 0 Product

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Homework Help Overview

The original poster seeks to find two polynomials of degree 2 in Z6[x] whose product results in a polynomial of degree 0. They also inquire whether the same can be achieved in Z7[x]. The context involves understanding polynomial degrees and the properties of polynomials over finite fields.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss specific polynomial examples and their degrees, questioning the conditions under which the product of two degree 2 polynomials can yield a degree 0 polynomial. There is also exploration of the implications of working in Z6 versus Z7, particularly regarding zero divisors.

Discussion Status

Participants have raised relevant questions and provided insights about the nature of zero divisors in Z6 and Z7. The discussion is ongoing, with some clarity achieved regarding the differences between the two fields, but no consensus or resolution has been reached yet.

Contextual Notes

The original poster expresses confusion about the degree of the product of polynomials and the specific characteristics of Z6 and Z7, particularly regarding the presence of zero divisors in Z6.

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Homework Statement



Find two polynomials, each of degree 2, in Z6[x] whose product has degree 0. Can you repeat the same in Z7[x]? Explain.

Homework Equations



In Z6[x] and Z7[x] can the only variable be x?

The Attempt at a Solution



I know Z6 consists of {0,1,2,3,4,5} and Z7: {0,1,2,3,4,5,6}; and I have tried (x2+5)(x2-3) and others but I get degree of 4, and it must be degree of 6 for it to be degree 0 in Z6..Please help, am I confused on how to solve this.
 
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What's 3x^2 times 2x^2 in Z6?
 
6x^4, so 0x^4...the polynomial is of degree 4, but since the coefficient is 0, the product would have a degree of 0?
 
sarah77 said:
6x^4, so 0x^4...the polynomial is of degree 4, but since the coefficient is 0, the product would have a degree of 0?

Sure. Now why can't that happen in Z7?
 
Since it is a prime number, no two elements in Z7 can be multiplied to obtain a number divisible by 7.
 
sarah77 said:
Since it is a prime number, no two elements in Z7 can be multiplied to obtain a number divisible by 7.

Exactly. There are no zero divisors in Z7. There are in Z6.
 
Thank you, that makes sense!
 

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