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Math Help - Reducible and irreducible polynomials.

  1. #1
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    Reducible and irreducible polynomials.

    Hey guys.

    Alright. I have this three-part assignment, but im really stuck on the second question. So if you could give me some pointers, it would be nice.

    Let R = F_3 [X] (Where the F is the double-stroke F. Didnt know how to do this.)

    (i) Show that X^2+1, X^2+X+1 and X^2+2X+2 are the only monic irreducible polynomials of degree 2 in R.

    I have already solved this question. I just listed all monic polynomials of degree 2 in R, and made a table showing which functions had roots and which where irreducible.

    (ii) Show that if a polynomial f \in R of degree 4 or 5 with no roots is reducible, then there is a monic irreducible polynomial of degree 2 dividing f

    This is the thing I dont know how to prove.
    A section of my textbook says that a polynomial ring (in this case R) is a unique factorization domain, which means that if f \in R isn't irreducible, there is a factorization f = f_1 f_2 such that 0 < deg(f_1) , deg(f_2) < deg(f) and deg(f) = deg(f_1) + deg(f_2).
    I also have a part that says that if R is a domain (which I believe is the case here), we have that V(fg) = V(f) \cup V(g) where V(x) is defined as the set of roots of x.
    So we can say, that there exists f_1 and f_2, where both have no roots.
    I have a feeling, that's the way I'm supposed to go, but I can't quite figure out how to tie it all together. So if someone could give me a hint, or tell me if im going the wrong way?

    Thanks in advance.

    /Morten
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Reducible and irreducible polynomials.

    First, Lets start with Deg = 4 Case.

    Now say we have a polynomial P(x) such that deg(P(x)) = 4.
    Now if P(x) is reducible then there are f(x),g(x)  \in F[X] such P(x) = f(x)g(x).
    Now i say that Deg(f(x)) = Deg(g(x)) = 2 (Why? I'll let you take a moment to think about this, do ask if you want to know why.)

    So since a quadratic is reducible iff it has roots in the underlying field.
    This would mean, if f(x) was reducible, then  \exists a \in F such that  f(a) = 0, which would mean f(a)g(a) = P(a) = 0, since P has no roots in F, this cannot be and f(x) is irreducible. Same thing for g(x). So P(x) = f(x)g(x) where f,g are irreducible, f is deg 2 (The monic part is easy, just divide out by the leading coeff of f(x) or g(x).

    Now for Deg(P(x)) = 5, It is the same, if P(x) = f(x)g(x) then Deg(f(x)) = 3 and Deg(g(x)) = 2, or Deg(f(x)) = 2 and Deg(g(x) = 3 (Again why?). For brevity Say Deg(f(x)) = 3, Deg(g(x))=2, Now, again using the same thing, g(x) reduces iff roots exist and if roots existed, then P(x) would also have roots, which cannot be so, g(x) is irreducible,
    P(x) = f(x)g(x).
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  3. #3
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    Re: Reducible and irreducible polynomials.

    Thanks for the help.
    Why do you say the monic part is simple?
    If I say a_2 X^2 + a_1 X + a_0 = a_2(X^2 + a_1/a_2 X + a_0/a_2) how can I be sure that a_1/a_2 \in Z and a_0/a_2 \in Z?
    Thats only valid if a_2 divides a_1 and a_0.
    Or have a misunderstood what kind of elements are inside F_3?
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  4. #4
    Super Member ILikeSerena's Avatar
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    Re: Reducible and irreducible polynomials.

    Hi m112358!

    F_3 or \mathbb{F}_3 or \mathbb{Z}/3 \mathbb{Z} is the field of the set {0,1,2} with the operations + and \times modulo 3.

    Since the calculations are modulo 3, every division is well defined, except divisions by 0.

    Btw, to get the black board F, use \mathbb{F}.
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  5. #5
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    Re: Reducible and irreducible polynomials.

    Arh... So 1/2 \equiv 1/(-1) = -1 \equiv 2 because it's modulo.

    Now I understand everything, thanks both of you Felt a bit stupid on the last bit there.

    /Morten
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  6. #6
    Senior Member jakncoke's Avatar
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    Re: Reducible and irreducible polynomials.

    Yes, Since the leading coefficient if its not 1, is certainly not 0, you divide by non zero element of  F , since a field has units, division is possible.
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    Re: Reducible and irreducible polynomials.

    Alright. I dont know if I should put this question in its own forum post, but now I just add it here, because it's related to the other questions.

    As you might have noticed, I mentioned that it was a three-part problem.
    I've been looking at the third part, but I'm getting stuck here as well.

    (iii) Show that X^5 - X + 1 is an irreducible polynomial in R and that L = R/<X^5 - X + 1> is a field with 243 elements. Let \alpha = [X]. Find an element \gamma \in L such that \alpha \gamma = 1 in L.

    Alright. I've looked at it a bit. First I'm looking at the irreducability of X^5 - X + 1. I have a proposition that says "The ideal <f> is a maximal ideal if and only if f is irreducible. In this case the quotient ring F[X]/<f> is a field.".
    So I'm guessing that first point on the agenda is proving that <X^5 - X + 1> is a maximal idea. But when I look through the stuff on maximal ideals in my book, all I find that seems usefull is "An ideal I \in R is maximal if and only if R/I is a field." But I have absolutely no idea on how to prove that R/I is a field.
    I'm guessing that I have to show, that if we assume there is an ideal J where I = <X^5 - X + 1> and I \subset J implies that J = R. So if we imagine an ideal J that contains I and elements not part of I. Then we have to prove that J = R. The only way I can think of, is by proving that 1 \in J.
    I just dont know how to do that. Any tips?
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    Re: Reducible and irreducible polynomials.

    I've read some proofs where they do something like this:

    Let I = <X^5 - X + 1>. Then we define J as an ideal containing I. Choose m \in J \backslash I, where deg(m) = 1. Then we have that m is not in I, so X^5-X+1 doesn't divide m, and since X^5-X+1 doesnt have any roots, m doesnt divide X^5-X+1 and therefore gcd(m, X^5 - X + 1) = 1.
    Then according to Euclid there exists s, t \in \mathbb{Z}, 1 = sm + t(X^5-X+1) and then we have that m \in J \Rightarrow sm \in J and X^5-X+1 \in I \subset J \Rightarrow t(X^5-X+1) \in J which together gives us that 1 \in J \Rightarrow J = R.

    Now I'm just thinking, couldn't this be done with every I? Is this really a proper proof that <X^5 - X + 1> is maximal?
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    Re: Reducible and irreducible polynomials.

    God I'm an idiot. I can use (ii) to show that X^5 - X + 1 is irreducible. I guess I'm getting a bit too tired for this. Thanks for the help guys

    /Morten
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  10. #10
    Senior Member jakncoke's Avatar
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    Re: Reducible and irreducible polynomials.

    Quote Originally Posted by m112358 View Post
    God I'm an idiot. I can use (ii) to show that X^5 - X + 1 is irreducible. I guess I'm getting a bit too tired for this. Thanks for the help guys

    /Morten
    Not only you, me 2, i just proved that and i compleltely forget about it. t.t
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  11. #11
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    Re: Reducible and irreducible polynomials.

    Quote Originally Posted by m112358 View Post
    Alright. I dont know if I should put this question in its own forum post, but now I just add it here, because it's related to the other questions.

    As you might have noticed, I mentioned that it was a three-part problem.
    I've been looking at the third part, but I'm getting stuck here as well.

    (iii) Show that X^5 - X + 1 is an irreducible polynomial in R and that L = R/<X^5 - X + 1> is a field with 243 elements. Let \alpha = [X]. Find an element \gamma \in L such that \alpha \gamma = 1 in L.

    Alright. I've looked at it a bit. First I'm looking at the irreducability of X^5 - X + 1. I have a proposition that says "The ideal <f> is a maximal ideal if and only if f is irreducible. In this case the quotient ring F[X]/<f> is a field.".
    So I'm guessing that first point on the agenda is proving that <X^5 - X + 1> is a maximal idea. But when I look through the stuff on maximal ideals in my book, all I find that seems usefull is "An ideal I \in R is maximal if and only if R/I is a field." But I have absolutely no idea on how to prove that R/I is a field.
    I'm guessing that I have to show, that if we assume there is an ideal J where I = <X^5 - X + 1> and I \subset J implies that J = R. So if we imagine an ideal J that contains I and elements not part of I. Then we have to prove that J = R. The only way I can think of, is by proving that 1 \in J.
    I just dont know how to do that. Any tips?
    no, the first thing to do is show that f(x) is irreducible. then <f(x)> is maximal. why? because F[x] for any field F is a principal ideal domain (why? because it is a euclidean domain with d(p(x)) = degree of p(x), and all euclidean domains are PIDs).

    that is, in F[x], for two ideals I and J, we have I = <p(x)> and J = <q(x)> for some polynomials p and q, so that I is contained in J only if q(x) divides p(x). but if f(x) is irreducible, the only proper factors it has are units, and for any unit u of any ring R, <u> = R, the entire ring. that is irreducible polynomials generate maximal ideals in F[x]. that's why you did parts (i) and (ii) first.

    and then, since <f(x)> is maximal, F[x]/<f(x)> is a field. we can just count the cosets to find out out big it is. since deg(x5+x+1) = 5, every coset can be written in the form:

    ax4+bx3+cx2+dx+e + <x5+x+1>, and since the difference of any 2 polynomials of degree 4 or less is also a polynomial of degree 4 or less, no two of these cosets can coincide (every element of the ideal generated by f is at least of degree 5). that gives us 3 choices for a, 3 choices for b, 3 choices for c, 3 choices for d, and 3 choices for e, for a total of 35 = 243 elements of F[x]/<f(x)> (243 distinct cosets).

    it is common to write the elements of F[x]/<f(x)> by writing x + <f(x)> = a, and considering the evaluation map F[x]-->F(a) (f(x)-->f(a)). this lets us write the elements of F[x]/<f(x)> as "polynomials in a", where we multiply "almost like normal", except for using:

    a5 = -a - 1, to keep all our expressions of degree 4 or less.

    note that -a5 - a = 1, that is: a(-a4 - 1) = 1, so the inverse of a is evidently -a4 - 1 (or 2a4 + 2, if you prefer, since (mod 3) -1 = 2).
    Last edited by Deveno; December 23rd 2012 at 05:15 AM.
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