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 (Where the F is the double-stroke F. Didnt know how to do this.)

**(i) Show that 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 of degree 4 or 5 with no roots is reducible, then there is a monic irreducible polynomial of degree 2 dividing **

This is the thing I dont know how to prove.

A section of my textbook says that a polynomial ring (in this case ) is a unique factorization domain, which means that if isn't irreducible, there is a factorization such that and .

I also have a part that says that if is a domain (which I believe is the case here), we have that where is defined as the set of roots of .

So we can say, that there exists and , 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

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) 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 such that , which would mean f(a)g(a) = P(a) = 0, since P has no roots in , 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).

Re: Reducible and irreducible polynomials.

Thanks for the help.

Why do you say the monic part is simple?

If I say how can I be sure that and ?

Thats only valid if divides and .

Or have a misunderstood what kind of elements are inside ?

Re: Reducible and irreducible polynomials.

Hi m112358! ;)

or or is the field of the set {0,1,2} with the operations and 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}.

Re: Reducible and irreducible polynomials.

Arh... So because it's modulo.

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

/Morten

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 , since a field has units, division is possible.

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 is an irreducible polynomial in and that is a field with 243 elements. Let . Find an element such that in .**

Alright. I've looked at it a bit. First I'm looking at the irreducability of . I have a proposition that says *"The ideal is a maximal ideal if and only if f is irreducible. In this case the quotient ring is a field."*.

So I'm guessing that first point on the agenda is proving that 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 is maximal if and only if is a field."* But I have absolutely no idea on how to prove that is a field.

I'm guessing that I have to show, that if we assume there is an ideal where and implies that . So if we imagine an ideal that contains and elements not part of . Then we have to prove that . The only way I can think of, is by proving that .

I just dont know how to do that. Any tips?

Re: Reducible and irreducible polynomials.

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

Let . Then we define as an ideal containing . Choose , where . Then we have that is not in , so doesn't divide m, and since doesnt have any roots, doesnt divide and therefore gcd(m, X^5 - X + 1) = 1.

Then according to Euclid there exists and then we have that and which together gives us that .

Now I'm just thinking, couldn't this be done with every I? Is this really a proper proof that is maximal?

Re: Reducible and irreducible polynomials.

God I'm an idiot. I can use **(ii)** to show that is irreducible. I guess I'm getting a bit too tired for this. Thanks for the help guys :)

/Morten

Re: Reducible and irreducible polynomials.

Quote:

Originally Posted by

**m112358** God I'm an idiot. I can use

**(ii)** to show that

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

Re: Reducible and irreducible polynomials.

Quote:

Originally Posted by

**m112358** 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 is an irreducible polynomial in and that is a field with 243 elements. Let . Find an element such that in .**
Alright. I've looked at it a bit. First I'm looking at the irreducability of

. I have a proposition that says

*"The ideal is a maximal ideal if and only if f is irreducible. In this case the quotient ring is a field."*.

So I'm guessing that first point on the agenda is proving that

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 is maximal if and only if is a field."* But I have absolutely no idea on how to prove that

is a field.

I'm guessing that I have to show, that if we assume there is an ideal

where

and

implies that

. So if we imagine an ideal

that contains

and elements not part of

. Then we have to prove that

. The only way I can think of, is by proving that

.

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(x^{5}+x+1) = 5, every coset can be written in the form:

ax^{4}+bx^{3}+cx^{2}+dx+e + <x^{5}+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 3^{5} = 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:

a^{5} = -a - 1, to keep all our expressions of degree 4 or less.

note that -a^{5} - a = 1, that is: a(-a^{4} - 1) = 1, so the inverse of a is evidently -a^{4} - 1 (or 2a^{4} + 2, if you prefer, since (mod 3) -1 = 2).