Abstract Algebra  Polynomials: Irreducibles and Unique Factorizationby VinnyCee Tags: abstract algebra, algebra, irreducible, polynomials 

#1
Apr711, 06:41 AM

P: 492

1. The problem statement, all variables and given/known data
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. 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) [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. 3. The attempt at a solution Obviously, x(x+1) is one of the ways to factor it, but what is the other? 



#2
Apr711, 07:55 AM

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P: 25,165

Try and find another root of f(x)=x^2+x besides 0 and 1. That will tell you how to factor it.




#3
Apr711, 02:57 PM

P: 492

Well, since it is in [tex]\mathbb{Z}_6[/tex], I know that 6 is congruent to 0 mod 6. But what is congruent to 1 mod 6? 1 is not even in [tex]\mathbb{Z}_6[/tex], right?
[tex]6\,\equiv\,0\,\left(mod\,6\right)[/tex] but [tex]???\,\equiv\,1\,\left(mod\,6\right)[/tex] 



#4
Apr711, 04:56 PM

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P: 16,543

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...




#5
Apr711, 05:09 PM

P: 492

Not relevant to the problem above, but... here is the long division for problem 4.4.2(a)!
[tex] \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{23} \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{23} & \multicolumn{2}{l}{\hspace*{72pt}x^9\,+\,x^8} & \multicolumn{2}{l}{\hspace*{98pt}\left(x^9\,\,x^8\right)} \\ \cline{23} & \multicolumn{2}{l}{\hspace*{108pt}2\,x^8} & \multicolumn{2}{l}{\hspace*{134pt}\left(2\,x^8\,\,2\,x^7\right)} \\ \cline{23} & \multicolumn{2}{l}{\hspace*{164pt}2\,x^7} & \multicolumn{2}{l}{\hspace*{190pt}\left(2\,x^7\,\,2\,x^6\right)} \\ \cline{23} & \multicolumn{2}{l}{\hspace*{220pt}2\,x^6} & \multicolumn{2}{l}{\hspace*{246pt}\left(2\,x^6\,\,2\,x^5\right)} \\ \cline{23} & \multicolumn{2}{l}{\hspace*{276pt}2\,x^5} & \multicolumn{2}{l}{\hspace*{302pt}\left(2\,x^5\,\,2\,x^4\right)} \\ \cline{23} & \multicolumn{2}{l}{\hspace*{332pt}2\,x^4} & \multicolumn{2}{l}{\hspace*{358pt}\left(2\,x^4\,\,2\,x^3\right)} \\ \cline{23} & \multicolumn{2}{l}{\hspace*{388pt}2\,x^3} & \multicolumn{2}{l}{\hspace*{414pt}\left(2\,x^3\,\,2\,x^2\right)} \\ \cline{23} & \multicolumn{2}{l}{\hspace*{444pt}2\,x^2} & \multicolumn{2}{l}{\hspace*{470pt}\left(2\,x^2\,\,2\,x\right)} \\ \cline{23} & \multicolumn{2}{l}{\hspace*{500pt}2\,x} & \multicolumn{2}{l}{\hspace*{526pt}\left(2\,x\,\,2\right)} \\ \cline{23} & \multicolumn{2}{l}{\hspace*{556pt}2} \end{array} \end{equation*} [/tex] It cuts off at the end, but I think you can do the last two lines:)) 



#6
Apr711, 05:15 PM

P: 492

How do I find another root? 



#7
Apr711, 05:19 PM

Mentor
P: 16,543

There are only 6 elements in [tex]\mathbb{Z}_6[/tex]. I guess it's not too hard to check them all and see whether they are roots?




#8
Apr711, 05:40 PM

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P: 25,165





#9
Apr711, 07:10 PM

P: 492

1 and 2 produce [tex]x^2\,\,3\,x\,+\,2[/tex] so that doesn't work.
3 and 4 produce [tex]x^2\,\,7\,x\,+12\,\longrightarrow\,x^2\,\,x[/tex] so that doesn't work. 1 and 3 produce [tex]x^2\,\,4\,x\,+3[/tex] so that doesn't work. 1 and 4 produce [tex]x^2\,\,5\,x\,+4[/tex] so that doesn't work. 2 and 3 produce [tex]x^2\,\,5\,x\,+6\,\longrightarrow\,x^2\,\,5\,x[/tex] so that doesn't work. 2 and 4 produce [tex]x^2\,\,6\,x\,+8\,\longrightarrow\,x^2\,+\,2[/tex] so that doesn't work. There are no other options that I can see! Ohhhh.... wait a minute! What about 3 and 4? [tex]\left(x\,\,3\right)\,\left(x\,+\,4\right)\,=\,x^2\,+\,4\,x\,\,3\,x\,\,12\,=\,x^2\,+\,x[/tex] 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? 



#10
Apr711, 07:57 PM

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