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

**(i) Show that $\displaystyle X^2+1, X^2+X+1 and X^2+2X+2$ 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 $\displaystyle f \in R$ of degree 4 or 5 with no roots is reducible, then there is a monic irreducible polynomial of degree 2 dividing $\displaystyle f$**

This is the thing I dont know how to prove.

A section of my textbook says that a polynomial ring (in this case $\displaystyle R$) is a unique factorization domain, which means that if $\displaystyle f \in R$ isn't irreducible, there is a factorization $\displaystyle f = f_1 f_2$ such that $\displaystyle 0 < deg(f_1) , deg(f_2) < deg(f)$ and $\displaystyle deg(f) = deg(f_1) + deg(f_2)$.

I also have a part that says that if $\displaystyle R$ is a domain (which I believe is the case here), we have that $\displaystyle V(fg) = V(f) \cup V(g)$ where $\displaystyle V(x)$ is defined as the set of roots of $\displaystyle x$.

So we can say, that there exists $\displaystyle f_1$ and $\displaystyle f_2$, 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

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

Re: Reducible and irreducible polynomials.

Thanks for the help.

Why do you say the monic part is simple?

If I say $\displaystyle a_2 X^2 + a_1 X + a_0 = a_2(X^2 + a_1/a_2 X + a_0/a_2)$ how can I be sure that $\displaystyle a_1/a_2 \in Z $ and $\displaystyle a_0/a_2 \in Z$?

Thats only valid if $\displaystyle a_2$ divides $\displaystyle a_1$ and $\displaystyle a_0$.

Or have a misunderstood what kind of elements are inside $\displaystyle F_3$?

Re: Reducible and irreducible polynomials.

Hi m112358! ;)

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

Re: Reducible and irreducible polynomials.

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

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

/Morten

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 $\displaystyle F $, 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 $\displaystyle X^5 - X + 1$ is an irreducible polynomial in $\displaystyle R$ and that $\displaystyle L = R/<X^5 - X + 1>$ is a field with 243 elements. Let $\displaystyle \alpha = [X]$. Find an element $\displaystyle \gamma \in L$ such that $\displaystyle \alpha \gamma = 1$ in $\displaystyle L$.**

Alright. I've looked at it a bit. First I'm looking at the irreducability of $\displaystyle X^5 - X + 1$. I have a proposition that says *"The ideal $\displaystyle <f>$ is a maximal ideal if and only if f is irreducible. In this case the quotient ring $\displaystyle F[X]/<f>$ is a field."*.

So I'm guessing that first point on the agenda is proving that $\displaystyle <X^5 - X + 1>$ 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 $\displaystyle I \in R$ is maximal if and only if $\displaystyle R/I$ is a field."* But I have absolutely no idea on how to prove that $\displaystyle R/I$ is a field.

I'm guessing that I have to show, that if we assume there is an ideal $\displaystyle J$ where $\displaystyle I = <X^5 - X + 1>$ and $\displaystyle I \subset J$ implies that $\displaystyle J = R$. So if we imagine an ideal $\displaystyle J$ that contains $\displaystyle I$ and elements not part of $\displaystyle I$. Then we have to prove that $\displaystyle J = R$. The only way I can think of, is by proving that $\displaystyle 1 \in J$.

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 $\displaystyle I = <X^5 - X + 1>$. Then we define $\displaystyle J$ as an ideal containing $\displaystyle I$. Choose $\displaystyle m \in J \backslash I$, where $\displaystyle deg(m) = 1$. Then we have that $\displaystyle m$ is not in $\displaystyle I$, so $\displaystyle X^5-X+1$ doesn't divide m, and since $\displaystyle X^5-X+1$ doesnt have any roots, $\displaystyle m$ doesnt divide $\displaystyle X^5-X+1$ and therefore gcd(m, X^5 - X + 1) = 1.

Then according to Euclid there exists $\displaystyle s, t \in \mathbb{Z}, 1 = sm + t(X^5-X+1)$ and then we have that $\displaystyle m \in J \Rightarrow sm \in J$ and $\displaystyle X^5-X+1 \in I \subset J \Rightarrow t(X^5-X+1) \in J$ which together gives us that $\displaystyle 1 \in J \Rightarrow J = R$.

Now I'm just thinking, couldn't this be done with every I? Is this really a proper proof that $\displaystyle <X^5 - X + 1>$ is maximal?

Re: Reducible and irreducible polynomials.

God I'm an idiot. I can use **(ii)** to show that $\displaystyle X^5 - X + 1$ 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 $\displaystyle X^5 - X + 1$ 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

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

Alright. I've looked at it a bit. First I'm looking at the irreducability of $\displaystyle X^5 - X + 1$. I have a proposition that says *"The ideal $\displaystyle <f>$ is a maximal ideal if and only if f is irreducible. In this case the quotient ring $\displaystyle F[X]/<f>$ is a field."*.

So I'm guessing that first point on the agenda is proving that $\displaystyle <X^5 - X + 1>$ 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 $\displaystyle I \in R$ is maximal if and only if $\displaystyle R/I$ is a field."* But I have absolutely no idea on how to prove that $\displaystyle R/I$ is a field.

I'm guessing that I have to show, that if we assume there is an ideal $\displaystyle J$ where $\displaystyle I = <X^5 - X + 1>$ and $\displaystyle I \subset J$ implies that $\displaystyle J = R$. So if we imagine an ideal $\displaystyle J$ that contains $\displaystyle I$ and elements not part of $\displaystyle I$. Then we have to prove that $\displaystyle J = R$. The only way I can think of, is by proving that $\displaystyle 1 \in J$.

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(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).

Re: Reducible and irreducible polynomials.

Hello. Sorry, but I was just wondering… How do you use (ii) to show that X^5-X+1 is irreducible?