# Abstract Algebra - Polynomials: Irreducibles and Unique Factorization

by VinnyCee
Tags: abstract algebra, algebra, irreducible, polynomials
 P: 492 1. The problem statement, all variables and given/known data Show that $$x^2\,+\,x$$ can be factored in two ways in $$\mathbb{Z}_6[x]$$ as the product of nonconstant polynomials that are not units. 2. Relevant 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) $\in$ 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 $p(x)|a_1(x)\,a_2(x)\,\cdots\,a_n(x)$, then p(x) divides at least one of the $a_i(x)$. 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 $\pm\,1$ are the only units. 3. The attempt at a solution Obviously, x(x+1) is one of the ways to factor it, but what is the other?
 Sci Advisor HW Helper Thanks P: 24,975 Try and find another root of f(x)=x^2+x besides 0 and -1. That will tell you how to factor it.
 P: 492 Well, since it is in $$\mathbb{Z}_6$$, I know that 6 is congruent to 0 mod 6. But what is congruent to -1 mod 6? -1 is not even in $$\mathbb{Z}_6$$, right? $$6\,\equiv\,0\,\left(mod\,6\right)$$ but $$???\,\equiv\,-1\,\left(mod\,6\right)$$
Emeritus
Thanks
PF Gold
P: 15,871

## Abstract Algebra - Polynomials: Irreducibles and Unique Factorization

5 is congruent to -1 in (mod 6). But you'll need to find another root except 5 and 0...
 P: 492 Not relevant to the problem above, but... here is the long division for problem 4.4.2(a)! $$\begin{equation*} \begin{array}{rc@{}c} & \multicolumn{2}{l}{\, \, \, x^9\,+\,x^8\,+\,2x^7\,+\,2x^6\,+\,2x^5\,+\,2x^4\,+\,2x^3\,+\,2x^2\,+\,2 x\,+\,2} \vspace*{0.12cm} \\ \cline{2-3} \multicolumn{1}{r}{x\,-\,1 \hspace*{-4.8pt}} & \multicolumn{1}{l}{ \hspace*{-5.6pt} \Big) \hspace*{20pt} x^{10}\,+\,x^8} \\ & \multicolumn{2}{l}{ -\left(x^{10}\,-\,x^9\right)} \\ \cline{2-3} & \multicolumn{2}{l}{\hspace*{72pt}x^9\,+\,x^8} & \multicolumn{2}{l}{\hspace*{98pt}-\left(x^9\,-\,x^8\right)} \\ \cline{2-3} & \multicolumn{2}{l}{\hspace*{108pt}2\,x^8} & \multicolumn{2}{l}{\hspace*{134pt}-\left(2\,x^8\,-\,2\,x^7\right)} \\ \cline{2-3} & \multicolumn{2}{l}{\hspace*{164pt}2\,x^7} & \multicolumn{2}{l}{\hspace*{190pt}-\left(2\,x^7\,-\,2\,x^6\right)} \\ \cline{2-3} & \multicolumn{2}{l}{\hspace*{220pt}2\,x^6} & \multicolumn{2}{l}{\hspace*{246pt}-\left(2\,x^6\,-\,2\,x^5\right)} \\ \cline{2-3} & \multicolumn{2}{l}{\hspace*{276pt}2\,x^5} & \multicolumn{2}{l}{\hspace*{302pt}-\left(2\,x^5\,-\,2\,x^4\right)} \\ \cline{2-3} & \multicolumn{2}{l}{\hspace*{332pt}2\,x^4} & \multicolumn{2}{l}{\hspace*{358pt}-\left(2\,x^4\,-\,2\,x^3\right)} \\ \cline{2-3} & \multicolumn{2}{l}{\hspace*{388pt}2\,x^3} & \multicolumn{2}{l}{\hspace*{414pt}-\left(2\,x^3\,-\,2\,x^2\right)} \\ \cline{2-3} & \multicolumn{2}{l}{\hspace*{444pt}2\,x^2} & \multicolumn{2}{l}{\hspace*{470pt}-\left(2\,x^2\,-\,2\,x\right)} \\ \cline{2-3} & \multicolumn{2}{l}{\hspace*{500pt}2\,x} & \multicolumn{2}{l}{\hspace*{526pt}-\left(2\,x\,-\,2\right)} \\ \cline{2-3} & \multicolumn{2}{l}{\hspace*{556pt}2} \end{array} \end{equation*}$$ It cuts off at the end, but I think you can do the last two lines:))
P: 492
 Quote by micromass 5 is congruent to -1 in (mod 6). But you'll need to find another root except 5 and 0...
So, $$5\,\equiv\,-1\,\left(mod\,6\right)$$...

How do I find another root?
 Emeritus Sci Advisor Thanks PF Gold P: 15,871 There are only 6 elements in $$\mathbb{Z}_6$$. I guess it's not too hard to check them all and see whether they are roots?
HW Helper
Thanks
P: 24,975
 Quote by micromass There are only 6 elements in $$\mathbb{Z}_6$$. I guess it's not too hard to check them all and see whether they are roots?
Yes, please. Check them all. You already know 0 and 5 work. So you don't need to check them. Now you have only four possibilities to check!
 P: 492 1 and 2 produce $$x^2\,-\,3\,x\,+\,2$$ so that doesn't work. 3 and 4 produce $$x^2\,-\,7\,x\,+12\,\longrightarrow\,x^2\,-\,x$$ so that doesn't work. 1 and 3 produce $$x^2\,-\,4\,x\,+3$$ so that doesn't work. 1 and 4 produce $$x^2\,-\,5\,x\,+4$$ so that doesn't work. 2 and 3 produce $$x^2\,-\,5\,x\,+6\,\longrightarrow\,x^2\,-\,5\,x$$ so that doesn't work. 2 and 4 produce $$x^2\,-\,6\,x\,+8\,\longrightarrow\,x^2\,+\,2$$ so that doesn't work. There are no other options that I can see! Ohhhh.... wait a minute! What about 3 and -4? $$\left(x\,-\,3\right)\,\left(x\,+\,4\right)\,=\,x^2\,+\,4\,x\,-\,3\,x\,-\,12\,=\,x^2\,+\,x$$ So 0 and 5 are roots and 3 and -4 are roots. Is that the correct answer to this problem or is there something more general I should be writing in order to "show" this is true?
HW Helper
Thanks
P: 24,975
 Quote by VinnyCee 1 and 2 produce $$x^2\,-\,3\,x\,+\,2$$ so that doesn't work. 3 and 4 produce $$x^2\,-\,7\,x\,+12\,\longrightarrow\,x^2\,-\,x$$ so that doesn't work. 1 and 3 produce $$x^2\,-\,4\,x\,+3$$ so that doesn't work. 1 and 4 produce $$x^2\,-\,5\,x\,+4$$ so that doesn't work. 2 and 3 produce $$x^2\,-\,5\,x\,+6\,\longrightarrow\,x^2\,-\,5\,x$$ so that doesn't work. 2 and 4 produce $$x^2\,-\,6\,x\,+8\,\longrightarrow\,x^2\,+\,2$$ so that doesn't work. There are no other options that I can see! Ohhhh.... wait a minute! What about 3 and -4? $$\left(x\,-\,3\right)\,\left(x\,+\,4\right)\,=\,x^2\,+\,4\,x\,-\,3\,x\,-\,12\,=\,x^2\,+\,x$$ So 0 and 5 are roots and 3 and -4 are roots. Is that the correct answer to this problem or is there something more general I should be writing in order to "show" this is true?
Sure. That's all you need. Nothing more general required. That's a different factorization than x*(x+1), yes? BTW you don't have to check them in pairs. If 2 is a root of x^2+x (which it is) then you know (x-2) is a factor. Then you can just divide to find the other factor.
Emeritus
 Quote by VinnyCee 2 and 3 produce $$x^2\,-\,5\,x\,+6\,\longrightarrow\,x^2\,-\,5\,x$$ so that doesn't work.
By the way, 2 and 3 does give you the right answer since $$x^2-5x=x^2+x$$. This follows since -5=1 (mod 6).