A polynomial P ∈ F[X] is irreducible in F[X] if deg P is greater than 0 and P is not divisible by any polynomial Q ∈ F[X] such that 0 is less than deg Q is less than deg P.

NOTE: See Definition 1, here, for definition of polynomials and Definition 4, here, for definition of deg of a polynomial.

Lemma 1:

If a polynomial D divides an irreducible polynomial P,

Then, either D is a constant or deg D = deg P.

Proof:

(1) Let P be an irreducible polynomial

(2) Let D be a polynomial that divides P.

(3) Assume that D is not a constant and that deg D is less than deg P.

(4) Then D cannot divide P [by Definition 1 above]

(5) So we have a contradiction and we must reject our assumption at step #3.

QED

Corollary 1.1: Every polynomial of degree 1 is irreducible

(1) Assume that there exists a polynomial D of degree 1 that is not irreducible.

(2) Then there exists a polynomial P that divides D where deg P is less than deg D and deg P is greater than 0.

(3) But this is impossible since deg D = 1 (either deg P = 0 or deg P ≥ 1).

(4) So we reject our assumption at step #1.

QED

Theorem 2: Euclid's Lemma for Polynomials

For polynomials S,T,U if S divides TU and S is relatively prime to T, then S divides U.

Proof:

(1) Since GCD(S,T)=1, there exists polynomials V,W such that [See Corollary 3.1, here]:

SV + TW = 1

(2) Multiplying both sides by U gives us:

S(UV) + (TU)W = U

(3) This proves that S divides U since S divides TU.

QED

Corollary 2.1: Generalized Euclid's Lemma for Polynomials

If a polynomial S divides a product of r factors and is relatively prime to the first r-1 factors, then it divides the last one.

Proof:

(1) Let T = the first r-1 factors.

(2) Let U = the last factor

(3) Then S divides TU and GCD(S,T)=1 so using Theorem 2 above, we can conclude that S divides U.

QED

Corollary 2.2: If a polynomial is divisible by pairwise relatively prime polynomials, then it is divisible by their product

Proof:

(1) Let P

_{1}, ..., P

_{r}be relatively prime polynomials which divide a polynomial P.

(2) For r=1, P

_{1}= P so the corollary is true.

(3) Assume that it is true up to r-1.

(4) So, then we have:

P = P

_{1}*..*P

_{r-1}*Q

(5) By Corollary 2.1 above, since P

_{r}divides P, it follows that P

_{r}must divide Q.

(6) So, then it follows that P

_{1}...*...*P

_{r}must divide P.

QED

Theorem 3: Every non constant polynomial P ∈ F[X] is a finite product:

P = cP

_{1}P

_{2}...P

_{n}

where c ∈ F

^{x}and P

_{1}, P

_{2}, ..., P

_{n}are monic irreducible polynomials where this factorization is unique. Order of these polynomials is not unique and there may include repeats of the same polynomial more than once.

Proof:

(1) If P is irreducible, then P = c*P

_{1}where c is the leading coefficient of P and P

_{1}= c

^{-1}P is irreducible and monic. [See Definition 1 above and Theorem 2, here]

(2) If P is reducible, then it can be written as a product of two polynomials of degree less than deg P.

(3) If the two polynomials less than P are irreducible, then we are done. If not, then we can continue to break each reducible polynomial into a product of reducible polynomials and then using step #1, we break this product into the desired result.

(4) The next step is to prove that this product independent of order is unique.

(5) Assume that:

P = cP

_{1}*...*P

_{n}= dQ

_{1}*...*Q

_{m}

where c,d ∈ F

^{x}and P

_{1}, ..., P

_{n}, Q

_{1}, ..., Q

_{m}are irreducible polynomials.

(6) c = d since both equal the leading coefficient of P.

(7) So, we have:

P

_{1}*...*P

_{n}= Q

_{1}*...*Q

_{m}

(8) Using Corollary 2.1 above, we know that P

_{1}cannot be relatively prime to all Q

_{i}

(9) So, let's assume that there exists a Q

_{i}such that GCD(P

_{1},Q

_{i}) ≠ 1. Let's label it Q

_{1}

(10) Let D be the unique monic GCD of P

_{1},Q

_{1}. [See Theorem 2, here]

(11) Since P

_{1}is irreducible, then P

_{1}= cD for some c.

(12) Since P

_{1}is monic, then c = 1 and P

_{1}= D.

(13) Since Q

_{1}is monic and irreducible, we can use the same argument as in steps #11 and steps #12 to establish that Q

_{1}= D.

(14) Therefore, it follows that Q

_{1}= P

_{1}

(15) We can make the same argument for all n elements of P

_{i}so that we have P

_{2}= Q

_{2}, P

_{3}= Q

_{3}, ..., P

_{n}= Q

_{n}

(16) Since there must be a matched element for each P

_{i}, it follows that n ≤ m.

(17) But we can also make the same line of argument for Q

_{j}so that we also know that m ≤ n

(18) Therefore n = m.

QED

References

- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2001

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