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

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To find two degree 2 polynomials in Z6[x] whose product results in a degree 0 polynomial, one can multiply polynomials that yield a coefficient of 0. For example, multiplying (3x^2) and (2x^2) in Z6 gives 6x^4, which simplifies to 0x^4, resulting in a degree 0 polynomial. In contrast, Z7[x] does not allow for such products because it is a field with no zero divisors; thus, no two non-zero elements can multiply to produce zero. This distinction highlights the fundamental difference between polynomial behavior in Z6 and Z7. Understanding these properties is crucial for solving polynomial equations in different modular arithmetic systems.
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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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