Reducible and irreducible polynomials.

**m112358** · December 22nd 2012, 03:33 PM

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

**jakncoke** · December 22nd 2012, 04:05 PM

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).

**m112358** · December 22nd 2012, 04:15 PM

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 $F_3$ ?

**ILikeSerena** · December 22nd 2012, 04:25 PM

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}.

**m112358** · December 22nd 2012, 04:30 PM

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

**jakncoke** · December 22nd 2012, 04:35 PM

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.

**m112358** · December 22nd 2012, 05:38 PM

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?

**m112358** · December 22nd 2012, 06:04 PM

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?

**m112358** · December 22nd 2012, 06:21 PM

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

**jakncoke** · December 22nd 2012, 06:33 PM

Originally Posted by m112358

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

**Deveno** · December 23rd 2012, 05:11 AM

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 $X^5 - X + 1$ 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).

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