Question

A nonzero polynomial f(x) is said to be monic if the leading coefficient is equal to 1. Prove that every monic polynomial of positive degree can be written as a prduct of monic irreducible polynomials.
Please help

Answer 1

i wonder if this can be done with induction

Answer 2

\[P_1(x)=x-r_1 \text{ is an irreducible poylnomial }\]
\[P_2(x)=x^2-(r_1+r_2)x+r_1r_2=(x-r_1)(x-r_2) \text{ any 2nd } \\ \text{ degree monic polynomial of degree 2 can be factored } \\ \text{ as a product of monic irreducible polynomials }\]
Now suppose for some integer k we have
\[P_k(x)=x^k-(r_1+r_2+ \cdots +r_k)x^{k-1}\\ +(r_1r_2+r_1r_3+\cdots r_1r_k+\cdots r_{k-1}r_k)x^{k-2} \\ -(r_1r_2r_3 + \cdots r_{k-2}r_{k-1}r_k)x^{k-3} \cdots +r_1r_2 \cdots r_k \\ =(x-r_1)(x-r_2) \cdots (x-r_k)\]
then show \[P_{k+1}=(x-r_1) \cdots (x-r_{k+1})\]

Answer 3

Just as a clarification on the problem, are these polynomials over a field? Like the reals or complex numbers?

Answer 4

I was thinking it has to be the over the complex field

Answer 5

that is the way any polynomial can be factorable

Answer 6

Yeah, I believe it would need to be over a field.

Answer 7

f(x) can be on Q, R or C

Answer 8

or any field

Answer 9

Well when I think have C
I think that includes all subsets of C such as R and so on...

Answer 10

but not all subsets of C are a field.

Answer 11

that is true

Answer 12

How to define P(x2) like this? why can't it be x^2 + ax +b as a general form?

Answer 13

I think you have to drop the monic condition if you dont have a field. Since you can always divide by the leading coefficient that way.

Answer 14

One more question: Is it not that any polynomial can be reduce to irreducible polynomials with its roots?? if the coefficient of the given one is 1, so as its factors, am I right?

Answer 15

If you start with a general 2nd degree polynomial: $$x^2+ax+b,$$then you could say something like, "If its irreducible, we are done, if not it splits into: $$x^2+ax+b=f(x)g(x)$$ where the deg(f),deg(g)<2. Then by the inductive hypothesis f and g are irreducible.

Answer 16

For example, P_5 has at most 5 roots, so that P_5 can be reduced tor (ax -r1)(bx-r2)....
and if the leading coefficient of given one is 1, it means a*b*.... =1 iff a=b=c=...=1?

Answer 17

over an arbitrary field, we aren't always going to be able to factor a polynomial into 1st degree terms with the roots. For example: $$x^2+1$$ is irreducible over Q[x] and R[x]. $$x^2-2$$ is irreducible over Q[x].

Answer 18

does P5 have 5 roots (not just at most)
are we not looking at the complex numbers?

Answer 19

at most because sometimes, it has double root. And the max number of roots is 5

Answer 20

i mean P5 can have the zero 0 with multiplicty 2 but that still counts as 2 zeros

Answer 21

But it doesn't matter, just terminology.

Answer 22

but any P5 with coefficient of x^5 as 1 can be written as (x-r1)(x-r_2)...(x-r_5)

Answer 23

if the leading coefficient is not 1, like 2x^2 -4 , can not be written as (1x - something) (1x- somethingelse), and both them have the leading 1.

Answer 24

but your thing defined monic polynomials as having the coefficient of the term with highest exponents as 1
so why are we talking about anything else?

Answer 25

Should we be writing this proof as if we are in the complex numbers? Because the complex numbers have the property that every polynomial of degree n has exactly n roots (counting multiplicities), which makes this problem easier than it should be.

If the problem is, "Prove that every monic polynomial over a field can be written as a product of irreducible monic polynomials", then I dont think we should use the factorization (x-r1)(x-r2)... since we may not have that.

Answer 26

first sentence actually defines it @Loser66

Answer 27

@nerdguy2535 no, no, we have to prove it in any field, not just complex

Answer 28

@frecles we are talking about the monic and the irreducible polys

Answer 29

If we need to prove it every field I don't see how that is possible because x^2+1 cannot be factored over the reals or rationals like nerd said

Answer 30

x^2+1 is irreducible and monic, which is all thats required.

Answer 31

@Loser66 I know in your first sentence it tells we are only talking about polynomials where the leading term has coefficient one so I don't know why we are talking about 2x^2-4

Answer 32

@freckles So?? if x^2 +1 is irreducible poly in R, and its leading coefficient is 1, then it is good enough, right?

Answer 33

right but why are we talking 2x^2-4

Answer 34

i do x^2+1 is irreducible in the reals and rationals sorry on that one

Answer 35

@freckles I give a counterexample if the coefficient is not 1, hihihihi..

Answer 36

a counterexample to what?

Answer 37

counterexample if the coefficient of the given poly is not 1, we cannot have the coefficients of its factors are 1.

Answer 38

but we are only considering if it is 1

Answer 39

or so i thought

Answer 40

Yes, yes, I got you!!

Answer 41

What definition are you using for irreducibility?

Answer 42

Like, a polynomial is irreducible if and only if....
otherwise, its reducible.

Answer 43

Ok, let me type it:
We say that a polynomial in K[x] is irreducible if its degree is positive and it cannot be written as a product of 2 polynomials each of strictly smaller (positive) degree
That is the definition on my book.

Answer 44

That way we have something to start with. I think the idea of an inductive proof is great. Its just like how you prove that the integers can be factored into primes. I think we want to mimic that somehow.

Answer 45

So lets try inducting on the degree of the polynomial.

Answer 46

Our inductive statement is, "Any monic polynomial of degree n (or less) can be factored into monic irreducible polynomials". Do you think this is fine?

Answer 47

I guess I'm thinking Strong Induction since I put the (or less) in there.

Answer 48

For the base case, if n=1, we have a monic polynomial of degree 1, which by your definition is clearly irreducible.

Answer 49

Now we assume the statement holds for all polynomials of degree n or less, and try to show we can do the same for a monic degree n+1 polynomial.

If the monic degree n+1 polynomial is already irreducible, we are done.

If not, then by your definition is factors into a product of 2 polynomials with strictly less degree. Since the degree of our starting polynomial was n+1, the degrees of these two factors must be n or less. So we can use our inductive hypothesis here.

Answer 50

So these two factor can each be written in terms of monic irreducible polynomials, and by putting the two irreudcible monic factorizations together, you get a monic irreducible factorization for the degree n+1 polynomial.

Answer 51

I got your induction proof. Thanks a lot.
One more question: So that we assume that the factors are monic? Don't we have to prove they are monic?

Answer 52

Ah, yeah, that could be a problem >.< How do we get around that?

Answer 53

In my head i was thinking: $$c_nx^n+c_{n-1}x^{n-1}+\cdots +c_1x+c_0$$$$=c_n\left(x^n+\frac{c_{n-1}}{c_n}x^{n-1}+\cdots +\frac{c_{1}}{c_n}x+\frac{c_{0}}{c_n}\right)$$

Answer 54

Since I have a proposition. It says:
Any polynomials in K[x] of positive degree can be written as a product of irreducible polynomials
with the proof follows
So that I think the problem here is "monic"

Answer 55

Ah yes, factoring out the leading coefficient will work. You know the product of the leading coefficients in your two factors is 1. So factor them out and multiply them to make them go away.

Answer 56

@freckles Now you know why I give 2x^2-4 there, right?

Answer 57

"
A nonzero polynomial f(x) is said to be monic if the leading coefficient is equal to 1. Prove that every monic polynomial of positive degree can be written as a prduct of monic irreducible polynomials.
"

I'm still having a problem with this.
It says prove for every poly of pos deg can be written as a product of monic polys.
I was thinking this has to be strickly over the complex field.
Because x^2+1 is a monic poly of pos deg but cannot be written as a product of monic polys over any other field but the complex field

Answer 58

but I guess we could also say the same for x-1
which cannot be written as a product of monic polys over any field

Answer 59

Why?? x^2 +1 is an irreducible one,( in R, Q) nothing to do with it. in C , we follow your proof

Answer 60

I think I'm just lost

Answer 61

I think you confuse yourself. :)

Answer 62

In R, Q , x^2 +1 is monic irreducible, it is done.

Answer 63

So we are trying to prove:
If the monic poly is reducible, then it can be written as a product of irrducible monic polynomials.

Answer 64

I think the word "product" might by throwing you off. The polynomial x^2+1 can be considered as a product only involving one term.

Answer 65

in C, x^2+1 can be written as (x+i)(x-i), both them are monic and irreducible. Proof done in 1 step

Answer 66

but it is 1*(x^2+1)
and 1 is not a monic polynomial

Answer 67

ye, 1 is polynomial degree 0

Answer 68

Ok I think I get it

Answer 69

This is just like the problem, prove any natural number can be written as a product of primes. 3 is a "product of primes"

Answer 70

ok, back to my problem , please. How to prove the factors are monic?

Answer 71

Say they aren't monic:$$f(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$$ $$g(x)=b_nx^n+b_{n-1}x^{n-1}+\cdots+b_1x+b_0$$but you know that: $$c_n\cdot b_n=1$$ since their product is monic (we started with a monic polynomial).

Answer 72

nope

Answer 73

"nope" tp cn times bn being 1?

Answer 74

to*

Answer 75

the degree of f and g are not good.

Answer 76

aha ahaha, yes, you are right

Answer 77

arbitrary degrees, my bad my bad

Answer 78

ugh now I have to retype that lol >.<

Answer 79

hihihi just copy and paste

Answer 80

so say f and g are like: $$f(x)=c_kx^k+c_{k-1}x^{k-1}+\cdots+c_1x+c_0$$ $$f(x)=b_rx^r+b_{r-1}x^{r-1}+\cdots+b_1x+b_0$$ where k+r=n+1, and k and r are at least 1.

Answer 81

i wonder if the remainder theorem can be used here

Answer 82

like as far as is it irreducible not
and also
is it monic or not

Answer 83

We know $$c_k\cdot b_r=1$$ since their product is monic. Factor out the leading coefficients: $$f(x)=c_k\left(x^n+\frac{c_{n-1}}{c_k}x^{n-1}+\cdots+\frac{c_1}{c_k}x+\frac{c_0}{c_k}\right)$$ do the same for g.

Answer 84

then you will have: $$c_k\cdot \left(\text{monic polynomial}\right)b_r\cdot \left(\text{monic polynomial}\right)$$and you can cancel out the ck and br

Answer 85

oops, my degree is still wrong in that last post >.< Im sure you know whats going on though.

Answer 86

I got you. Thanks a lot for the help. @nerdguy2535 and @freckles

Answer 87

I think nerd helped you way more. I probably just confused everyone more. :p

Answer 88

Good question loser.

Answer 89

If you are interesting, you are welcome at my problems. hihihihi.... To me, it 's not good at all. I got stuck and not many people here can help. I do appreciate if someone makes my question clear or gives me another perspective on it. It helps me understand the concept deeper.