Abstract Algebra - Polynomials: Irreducibles and Unique Factorization

In summary, we have shown 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: x*(x+1) and (x-3)*(x+4).
  • #1
VinnyCee
489
0

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?
 
Last edited:
Physics news on Phys.org
  • #2
Try and find another root of f(x)=x^2+x besides 0 and -1. That will tell you how to factor it.
 
  • #3
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
5 is congruent to -1 in (mod 6). But you'll need to find another root except 5 and 0...
 
  • #5
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\,+\,2x\,+\,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*} [/tex]It cuts off at the end, but I think you can do the last two lines:))
 
  • #6
micromass said:
5 is congruent to -1 in (mod 6). But you'll need to find another root except 5 and 0...

So, [tex]5\,\equiv\,-1\,\left(mod\,6\right)[/tex]...

How do I find another root?
 
  • #7
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
micromass said:
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?

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!
 
  • #9
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?
 
Last edited:
  • #10
VinnyCee said:
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?

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.
 
  • #11
VinnyCee said:
2 and 3 produce [tex]x^2\,-\,5\,x\,+6\,\longrightarrow\,x^2\,-\,5\,x[/tex] so that doesn't work.

By the way, 2 and 3 does give you the right answer since [tex]x^2-5x=x^2+x[/tex]. This follows since -5=1 (mod 6).
 

1. What is abstract algebra?

Abstract algebra is a branch of mathematics that studies algebraic structures such as groups, rings, and fields. It focuses on the properties and operations of these structures rather than specific numbers or equations.

2. What are polynomials?

Polynomials are mathematical expressions that consist of variables and coefficients, and are made up of terms that are combined using addition, subtraction, and multiplication. Examples of polynomials include x^2 + 3x + 2 and 5x^3 - x + 7.

3. What does it mean for a polynomial to be irreducible?

A polynomial is irreducible if it cannot be factored into polynomials of lower degree with coefficients in the same field. In other words, it cannot be broken down into simpler parts.

4. What is unique factorization in abstract algebra?

Unique factorization is the property of certain algebraic structures, such as polynomials, where every element can be expressed as a unique product of irreducible elements. This means that there is only one way to break down a polynomial into its irreducible factors.

5. Why is unique factorization important in abstract algebra?

Unique factorization allows us to solve equations and manipulate polynomials efficiently, without having to consider multiple possible factorizations. It also helps us understand the structure of algebraic systems and prove theorems about them.

Similar threads

  • Calculus and Beyond Homework Help
Replies
9
Views
981
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
981
  • Calculus and Beyond Homework Help
Replies
17
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
411
  • Linear and Abstract Algebra
Replies
5
Views
877
Back
Top