6. PID and UFD Let R be a commutative ring. Recall that a non-unit x R is called irreducible if x cannot be written as a product of two non-unit elements of R i.e.∈x = ab implies either a is an unit or b is an unit. Also recall that a domain R is called a principal ideal domain or a PID if every ideal in R can be generated by one element, i.e. is principal. 6.1. Lemma. (a) Let R be a commutative domain. Then prime elements in R are irreducible. (b) Let R be a PID. Then an irreducible in R is a prime element. Proof. (a) Let (p) be a prime ideal in R. If possible suppose p = uv.Thenuv (p), so either u (p)orv (p), if u (p), then u = cp,socv = 1, that is v is an unit. Similarly,∈ if v (p), then∈ u is an∈ unit. ∈ ∈(b) Let p R be irreducible. Suppose ab (p). Since R is a PID, the ideal (a, p)hasa generator, say∈ x, that is, (x)=(a, p). Then ∈p (x), so p = xu for some u R. Since p is irreducible, either u or x must be an unit and we∈ consider these two cases seperately:∈ In the first case, when u is an unit, then x = u−1p,soa (x) (p), that is, p divides a.Inthe second case, when x is a unit, then (a, p)=(1).So(∈ ab,⊆ pb)=(b). But (ab, pb) (p). So (b) (p), that is p divides b. ⊆ ! ⊆ Because of the above result, we shall not distinguish between prime elements and irre- ducible elements in a PID. 6.2.ExamplesofPIDs:Let k be a field. Then we have seen k[x] is a PID. Also Z is a PID. A little later, we shall prove that Z[i]andZ[e2πi/3] are examples of PID. We have also seen in homework exercise that Z[√ 5] is not a PID. Also one easily see that k[x, y] is not a PID, since one can prove that the− ideal (x, y) cannot be generated by a single element. 6.3. Definition. Let R be a ring and a , ,a R. Then the ideal (a , ,a )hasa 1 ··· n ∈ 1 ··· n generator x. Verify that x divides each aj and if y is some element of R that divides each a ,theny divides x.Sowesaythatx is the g.c.d. of a , ,a . Since the generator of a j 1 ··· n principal ideal is unique up to units, the gcd of a1, ,an is unique upto units. We say that two elements a and b in a PID are relatively prime,··· if their g.c.d. is 1. We repeat that when we say d is the greatest common divisor of a1 and a2, it simply means that d is a generator for the ideal (a1,a2), that is, (d)=dR =(a1,a2). So the gcd d can be written in the form d = ra + sa for some r, s R. 1 2 ∈ 6.4. Definition. A commutative ring R is called Noetherian if given any increasing sequence of ideals I I I in R, there exists a natural number n such that I = I = 1 ⊆ 2 ⊆ 3 ⊆ ··· n n+1 In+2 = .Inwords,wesaythatR is Noetherian if any increasing sequence of ideals in R stabilizes.··· 6.5. Lemma. APIDisNoetherian. Proof. Let R be a PID. Let (a ) (a ) (a ) be a increasing sequence of ideals in 1 ⊆ 2 ⊆ 3 ⊆ ··· R. Then verify that I = (ai) is an ideal, so there exists a R such that I =(a). There is a j 1suchthata (a ∪). It follows that (a)=(a )=(a ∈)= . ! ≥ ∈ j j j+1 ··· 6.6. Definition. A domain R is called an unique factorization domain or an UFD if every nonzero element can be written, uniquely upto units as a product of irreducible elements. 6.7. Theorem. Every PID is an UFD. 16 Proof. Fix a a R. We want to write a as a product of primes (equivalently irreducibles) and show that∈ such a decomposition is unique upto permutation of the prime factors and upto units. Step 1: Any non-unit a is divisible by an irreducible element. Suppose not. Since a is not irreducible write a = a b where a ,b are non-units. Since a a, a is not irreducible, so 1 1 1 1 1 | 1 write a1 = a2b2 where a1,b1 are non-units. Continuing this way we get a strictly increasing infinite sequence of ideals (a1) ! (a2) ! (a3) ! which, is not possible by lemma 6.5. This proves step 1. ··· Step 2: Any a is a product of irreducibles and an unit. Suppose not. By step 1, write a = p1c1 where p1 is an irreducible. Then c1 is not a unit. So write c1 = p2c2 where p2 is irreducible. Continuing this way we get a sequence (c1) ! (c2) ! (c3) ! which, is not possible by lemma 6.5. So This proves step 2. ··· Step 3: By step 2 we can write a = p p p where p are irreducible elements, not 1 2 ··· r i necessarily all distinct. Let a = p1p2 pr = q1 qs be two such decompositions. Each qj is a prime and q p p ,henceq p···for some ···i,henceq = u p for some unit u . Similarly j | 1 ··· r j | i j j i j each pi is equal to some qj upto a unit. If there are more p’s than q’s then canceling all the q’s will yield a product of p ’s equal to an unit which is impossible. So r = s and pi and qi are same upto units and upto permutation. ! 6.8. Remark. The rings Z, k[x], Z[i], Z[ω] are all UFD’s. This in particular proves that every integer can be written uniquely a a product of positive primes and 1andthatevery polynomial in one variable can be written as a product of irreducible polynomials± that are unique upto a scalar. Let R be an UFD and a R. We can write a = pe(p) where the product is over distinct ∈ p primes of R and almost all e(p) is zero. The numbers! e(p) is uniquely determined by a and p.Infacte(p) is the largest integer n such that pn a. This is because a′ = a/pe(p) is a ′ e(p|)+1 product over primes different from p,sop " a , i.e. p " a. We write e(p)=ordp(a). 6.9. Lemma (Exercise). Let R be a PID and a, b, c R.Suppose(a, b)=(1)and a bc. Then a c. ∈ | | Proof. Since (a, b) = (1), there exists r, t R such that ar + bt =1.Soc = acr + bct. Since a divides both acr and bct, it follows that∈a divides c. ! 6.10.ExerciseLet I, J be two ideals in a commutative ring R.Letx , ,x be a 1 ··· m set of generators for I and let y1, ,yn be a set of generators for J.Thenshowthat x y : i =1, ,m,j =1, ,n is··· a set of generators for the ideal IJ. { i j ··· ··· } 6.11.Exercise:Let R = Z[√ 5] = a + b√ 5: a, b Z . (a) Let I =(2, 1+√ 5) (recall− { that this− denotes∈ the} ideal in R generated by 2 and (1 + √ 5)). Let J =(2, 1− √ 5). Show that I = J.Showthat2R = IJ = I2. (b) Show− that R/I Z/−2Z −and conclude that I is a maximal ideal in R, hence a prime ideal. ≃ (c) Show that I is not a principal ideal. (Hint: Let u, v R.Observethatifu divides v in R,then u 2 divides v 2 in Z). ∈ | | | | 17 solution. (a) Show that 2 I and 1 √ 5=2 (1 + √ 5) I,soJ I. Again, 2 J ∈ − − − − ∈ ⊆ ∈ and (1 + √5) = 2 (1 √ 5) J,soI J.SoI = J. By the previous exercise, we know IJ =(2 2, 2(1+√− 5),−2(1 −√ ∈5), (1+√⊆5)(1 √ 5)) = (4, 2+2√ 5, 2 2√ 5, 6). Note that all· the four generators− − in the− ideal 2R−.soIJ− −2R.Ontheotherhand2=6− − − 4 IJ, so 2R IJ. It follows that I2 = IJ =2R. ⊆ − ∈ (b)⊆ If R = I, then we would have 1 = 1.1 I2, which would imply R = I2 =2R, but then 1 = 2x for some x R, which would∈ imply 1/2 R which is not true. So R = I. Also note that (1 + √∈ 5) I I2, since (1 + √ 5)∈/2 / R.SoI is a proper ideal̸ in R and I2 is a proper subset− ∈ of\I.NowR/I2 =−R/2R.∈ Verify that R/2R = 0+2R, 1+2R, √ 5+2R, 1+√ 5+2R and these four cosets are distinct, so R/2R =4. 2 { − − } R/I 2 | 2| By the third isomorphism theorem, we have R/I 2 . Since I = I ,wehave I/I > 1, ≃ I/I ̸ | | so R/I is a proper factor of 4, so R/I = 1 or 2. Finally R = I implies R/I =1,soR/I has| size| 2. Verify that the only commutative| | ring of size 2 is ̸Z/2Z.SoR/I| |Z̸ /2Z. (c) If z =(α + iβ) R, we write z 2 = zz¯ = α2 + β2 and call it the norm≃ of z.Note that the norm of every∈ element of R is| | an integer. Suppose u divides v in R.Thenu = vx for some x R.So¯u =¯vx¯,henceuu¯ = vvx¯ x¯ or u 2 = v 2 x 2.