Gauss s Theorem Theorem: Suppose R is a U.F.D.. Then R[x] is a U.F.D. To show this we need to constuct some discrete valuations of R. Proposition: Suppose R is a U.F.D. and that π is an irreducible element of R. Let K be the fraction field of R. Then there is a discrete valuation v π : K = K {0} Z defined by v π ( r s ) = ord π(r) ord π (s) for r, s R {0}, where r = c π ord π(r) r for some unit c R and some element r R which does not involve an associate of π in its factorization. 1
Proof of the Proposition: We check the properties of discrete valuations: 1. v π is surjective, since v π (π n ) = n. 2. v π (ab) = v π (a) + v π (b) using the factorizations of a and b. 3. v π (a + b) min(v π (a), v π (b)) if a + b 0. To check this, note that the left and right hand side both change by v π (s) if we multiply a and b by s R {0}. So we can reduce to the case in which a, b R. The right hand side is now the exponent of the largest power of π which divides both a and b in R, and this is lower bound for the power of π dividing a + b. 2
Example: When R = Z, the irreducibles have the form π = ±p for some prime p, and v π is the valuation v p we have seen before. Definition Suppose f = f(x) = a 0 + a 1 x + a n x n is a non-zero element of R[x]. irreducible element of R. Define Let π be an v π (f) = min{v π (a i ) : 0 i n, a i 0} The content of f is defined to be cont(f) = π v π(f) π S where S is a set of non-associate irreducible elements of R such that each irreducible is associate to exactly one element of S. Note that cont(f) is defined only up to multiplication by a unit. (It would be more natural to define cont(f) as an ideal of R.) 3
Gauss s Lemma: Suppose R is a U.F.D. and that f = f(x) and g = g(x). Then cont(fg) = cont(f) + cont(g) if we use the same set S of irreducibles to define cont(f) and cont(g). Proof: Suppose π S. It s enough to show v π (f g) = v π (f) + v π (g). (1) To prove this, note that we can divide f = f(x) by the constant π v π(f) to arrive at a polynomial f 1 = f 1 (x) = f(x) π v π(f) in R[x], where v π (f 1 ) = 0 since there is a non-zero coefficient of f 1 (x) which can t be further divided by π in R. Similarly, g 1 = g(x) π v π(g) R[x] and v π(g 1 ) = 0 4
Now we have So fg = π v π(f)+v π (g) f 1 g 1 v π (fg) = v π (f) + v π (g) + v π (f 1 g 1 ) and we want to show v π (f 1 g 1 ) = 0 This is equivalent to saying that not all coefficients of f 1 g 1 are divisible by π in R. Since v π (f 1 ) = 0, there is an integer i such that the coefficient a i of xi in f 1 = f 1 (x) is not divisible by π. Let i be the largest such integer. Similarly, let j be the largest integer such that the coefficient b j of xj in g 1 = g 1 (x) is not divisible by π. Expanding f 1 g 1 now shows the coefficient of x i+j in this product is a i b j mod π, and this is not divisible by π. So v π (f 1 g 1 ) = 0 and we re done. 5
Recall now that F is the fraction field of R. The ring F [x] is Euclidean (with respect to the norm which sends a non-zero polynomial to its degree). So F [x] is a P.I.D., and thus a U.F.D. So it makes sense to talk about irreducible elements of F [x]. Theorem: Suppose f(x) R[x] is irreducible in F [x] and cont(f(x)) = 1. Then f(x) is irreducible in R[x]. Proof: Fix the set S of representatives for the irreducibles of R. We want to show that if f(x) = g(x)h(x) (2) in R[x], then one of g(x) or h(x) is a unit in R[x]. 6
View (2) as a factorization in F [x]. Since f(x) is irreducible in F [x], one of g(x) or h(x) is a unit in F [x]. But the units of F [x] are just F = F {0}. So we can suppose g(x) = d is in F, by switching g(x) and h(x) if necessary. Gauss s lemma shows cont(f(x)) = cont(g(x)h(x)) = cont(g(x))cont(h(x)) where all of these contents are in R since f(x), g(x) and h(x) are in R. Since cont(f(x)) = 1, this proves cont(g(x)) = cont(d) is a unit in R. But since d F, the fact that R is a U.F.D. shows d = cont(d)u for some unit u R. Therefore d = g(x) is a unit in R, and we have shown f(x) is irreducible in R[x]. 7
Theorem: The ring R[x] is a U.F.D. Proof: Suppose f(x) R[x] is non-zero. In F [x], we have a factorization f(x) = c n i=1 p i (x) (3) in which c F [x] = F and p i (x) F [x] is irreducible in F [x]. Taking contents gives cont(f(x)) = cont(c) n i=1 cont(p i (x)) (4) Dividing the sides of (3) by those of (4) shows f(x) cont(f(x)) = c cont(c) n i=1 p i (x) cont(p i (x)) (5) 8
Now note that each of the ratios f(x) cont(f(x)), c p and i (x) cont(c) cont(p i (x)) have content 1. In particular, these polynomials lie in R[x]. The element R is a U.F.D.. c cont(c) has to be a unit in R since p i (x) We proved before that cont(p i (x)) = g i(x) is irreducible in R[x], since it has content 1 and is irreducible in F [x]. has a factor- So we ve shown each ization into irreducibles. f(x) cont(f(x)) Now factor cont(f(x)) in R as cont(f(x)) = u k z k where u R = R[x] and the z k are irreducible in R. The z k stay irreducible in R[x] since the only way to them would be as a product of constant polynomials. 9
This gives a factorization f(x) f(x) = cont(f(x)) cont(f(x)) = u z k p i (x) cont(p i (x)) k of f(x) into a product of irreducibles in R[x]. i (6) Suppose one had another factorization f(x) = u z s s j p j (x) (7) of f(x) into the product of a unit u of R[x], irreducible elements z s of R, and non-constant irreducible polynomials p j (x) R[x]. 10
We now view (6) and (7) in F [x] and use the fact that u, u, z k and z s are in F. Since the p i (x) were irreducible in F [x], for each i there is a j such that p j (x) = h i,j(x)p i (x) for some h i,j (x) F [x]. But then p j (x) cont(p j (x)) = h i,j (x) cont(h i,j (x)) p i (x) cont(p i (x)) in R[x]. Since p j (x) was assumed to be irreducible in R[x], and p i (x) is not a constant polynomial, this implies p j (x) = u p i (x) j cont(p i (x)) for some unit u j R[x] = R. We now cancel the terms corresponding to u j p i (x) cont(p i (x)) and p j (x) from the following two expressions for f(x): 11
u z k k i p i (x) cont(p i (x)) = u This leads to an equality u z k = u k s s z s z s p j (x) where u and u are units and the z k and z s are irreducibles in R. Since R is a U.F.D., we know that after rearranging the z s, there are exactly as many z s as there are z k, and z k is associate to z k for all k. This completes the proof of the essential uniqueness of the factorization of f(x) into a product of irreducibles in R[x]. Corollary of the Proof: The irreducible elements of R[x] are the irreducibles of R together with the non-constant polynomials h(x) R[x] which are irreducible in F [x] and have content 1. j 12
Irreducibility Criteria To factor elements of R[x] into irreducibles when R is a U.F.D., one needs to be able to recognize when an element of R[x] is irreducible. Proposition: Suppose R is a U.F.D. and that f(x) R is a monic polynomial (so the leading coefficient is 1). Suppose there is a proper ideal I of R such that the class [f(x)] in R[x]/(I R[x]) = (R/I)[x] is not the product of two polynomials of smaller degree in (R/I)[x]. Then f(x) is irreducible. Example: Suppose R = Z, I = Z 2 and f(x) = x 3 x 1. Since f(x) has not roots in Z/2, [f(x)] can have no linear factors in (Z/2)[x]. But since it has degree 3, [f(x)] would have a linear factor if it could be factored into a product of polynomials of lesser degree. Hence f(x) is irreducible in Z[x] 13
Proof: If f(x) is reducible, then f(x) = g(x) h(x) (8) for some g(x), h(x) R[x] neither of which are units. We have 1 = cont(f(x)) = cont(g(x)) cont(h(x)) by Gauss s Lemma. So cont(g(x)) and cont(h(x)) are units. Since we assumed g(x) and h(x) are not units, this implies neither g(x) nor h(x) is a constant polynomial. Hence (8) implies that both of g(x) and h(x) have degree less than the monic polynomial f(x). The reduction of (8) mod I is a factorization of f(x) mod I into polynomials of smaller degree, which we have supposed does not exist. This contradiction shows f(x) had to have been irreducible. 14
Warning: There are monic irreducible f(x) Z[x] whose image in (Z/p)[x] is reducible for every prime p. Class field Theory: Given a monic non-constant irreducible f(x) Z[x], describe how it factors in (Z/p)[x] for each prime p. For example, for which p is it a product of linear polynomials in (Z/p)[x]? The Cebotarev density theorem says that there will always be infinitely many p of this last kind. Example (from abelian class field theory): Suppose f(x) = x 2 + 1. If p = 2, this factors as (x + 1) 2. If p is odd, then x 2 + 1 factors into linear factors if and only if 1 is a square in (Z/p). We ll see later that (Z/p) is cylic of order (p 1)/2, and 1 has order 2 if p > 2. So x 2 + 1 factors into linear factors in (Z/p) if and only if p 1 mod 4. 15
Theorem: (Eisenstein s criterion) Suppose R is an integral domain and that P is a prime ideal of R. Suppose f(x) = x n + a n 1 x n 1 + + a 0 is a monic polynomial in R[x], and that a i P for all i < n, while a 0 P 2. Then f(x) is irreducible in R[x]. (One calls f(x) an Eisenstein polynomial with respect to P.) Example: When R = Z, the polynomial f(x) = x 5 + 2x + 2 is Eisenstein with respect to P = Z2. 16
Example: A non-constant polynomial h(x) is irreducible in R[x] if and only if h(x + a) is irreducible for all a R. This is because a factorization h(x + a) = t(x) s(x) would give a factorization h(x) = t(x a) s(x a). This can be used to convert some h(x) into Eisenstein polynomials h(x + a), thereby showing h(x) is irreducible. For example, suppose R = Z and p is prime. Let s show h(x) = x p 1 + x p 2 + x + 1 = xp 1 x 1 is irreducible. This follows from the fact that h(x + 1) = (x + 1)p 1 x = p k=1 ( p k ) x k 1 (9) is Eisenstein at Zp. This is because for 1 k < p, the binomial coefficient ( ) p k is congruent to 0 mod p, while ( ) p 1 = p 0 mod p 2 17
Proof of the Eisenstein criterion Suppose f(x) R[x] is Eisenstein with respect to the prime ideal P. Suppose f(x) is reducible, so that f(x) = a(x) b(x) with neither a(x) or b(x) a unit. Then neither a(x) or b(x) can be constant, since otherwise they would have to be units because f(x) is monic. Therefore a(x) and b(x) have degree less than n = deg(f(x)). Let f(x) be the image of f(x) in (R/P )[x]. We have f(x) = a(x) b(x). However, since f(x) is Eisenstein at P, all of its coefficients except for the leading one are in P. Since f(x) is also monic, this shows f(x) = x n = a(x) b(x) 18
Because P is a prime ideal, R/P is an integral domain, and R/P is a subring of the fraction field L of R/P. Thus we get an equality f(x) = x n = a(x) b(x) in the U.F.D. L[x]. Since x is an irreducible element of L[x], the only way this can happen is that a(x) = cx m and b(x) = c 1 x s (10) for some integers m, s 0 such that m + s = n and some non-zero element c of L. We showed earlier that a(x) and b(x) have degree less that n = deg(f(x)). So m, s < n, and it follows from m + s = n that m, s > 0. But now (10) implies the constant terms of a(x) and of b(x) are 0 mod P. The product of these constant terms is thus in P 2, but this is the constant term of f(x). This contradicts the assumption that f(x) is an Eisenstein polynomial. This contradiction shows f(x) had to have been irreducible in R[x]. 19