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VinnyCee
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Homework Statement
Show that [tex]x^2\,+\,x[/tex] can be factored in two ways in [tex]\mathbb{Z}_6[x][/tex] as the product of nonconstant polynomials that are not units.
Homework Equations
Theorem 4.8
Let R be an integral domain. then f(x) is a unit in R[x] if and only if f(x) is a constant polynomial that is a unit in R.Corollary 4.9
Let F be a field. then f(x) is a unit in F[x] if and only if f(x) is a nonzero constant polynomial.Definition
Let F be a field. A nonconstant polynomial p(x) [itex]\in[/itex] F[x] is said to be irreducible if its only divisors are its associates and the nonzero constant polynomials (units). A nonconstant polynomial that is not irreducible is said to be reducible.Theorem 4.10
Let f be a field. A nonzero polynomial f(x) is reducible in F[x] if and only if f(x) can be written as the product of two polynomials of lower degree.Theorem 4.11
Let F be a field and p(x) a nonconstant polynomial in F[x]. then the following conditions are equivalent:
(1) p(x) is irreducible.
(2) If b(x) and c(x) are any polynomials such that p(x)|b(x) c(x), then p(x)|b(x) or p(x)|c(x).
(3) If r(x) and s(x) are any polynomials such that p(x) = r(x) s(x), then r(x) or s(x) is a nonzero constant polynomial.Corollary 4.12
Let F be a field and p(x) an irreducible polynomial in F[x]. If [itex]p(x)|a_1(x)\,a_2(x)\,\cdots\,a_n(x)[/itex], then p(x) divides at least one of the [itex]a_i(x)[/itex].NOTE: An element a in a commutative ring with identity R is said to be an associate of an element b of R if a = bu for some unit u. In the integer ring, the only associate of an integer n are n and -n because [itex]\pm\,1[/itex] are the only units.
The Attempt at a Solution
Obviously, x(x+1) is one of the ways to factor it, but what is the other?
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