3 Factorisation into irreducibles

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1 3 Factorisation into irreducibles Consider the factorisation of a non-zero, non-invertible integer n as a product of primes: n = p 1 p t. If you insist that primes should be positive then, since n could be negative, you have to allow multiplication by a unit, n = u p 1 p t where the p i are primes and u is invertible (i.e. u = ±1). There are a number of points to note: existence of a prime decomposition (i.e. there s at least one way of splitting n as a product of primes); uniqueness of prime decomposition - of course you can write the primes in a different order but, since the ring is commutative, that s an inessential variation. You probably knew this fact (that each integer has just one prime decomposition) long before you saw a proof of it if, indeed, you have seen a proof; anyway, we ll prove it here (and in a more general context). The primes appearing can be multiplied by any unit, but that s also an inessential variation. For instance, we will count 3 as a prime and as being essentially the same prime as 3. In the case of the ring of integers we could insist that only positive integers can count as primes but, in more general cases, there may be no canonical choice like this, so we just have to accept that, e.g., 3 is as good as +3. In the terminology that we introduced in the previous section, 3 and 3 are associates (in particular they generate the same ideal). Allowing associates means we can always absorb the unit u (appearing above) in any of the primes in the decomposition of a non-zero non-unit. One important point which is rather hidden in the case of the ring of integers is that there are two concepts bundled together in the notion of prime integer. The first is irreducibility, meaning that the element can t be factorised in any non-trivial way; the second is primeness - if p divides a product then it divides one of the factors. In general these are different! There are two main sources of examples of the kinds of ring we consider in Section 3: rings of integers in number fields and polynomial rings. Number fields are what we get by taking the field Q of rationals and adding roots of polynomials; sitting inside such a field is an analogue of the integers: compare Z[ 2] sitting inside Q[ 2] with Z sitting inside Q (more generally, the ring of integers in the field Q[ d], where d is a square-free integer, is Z[ d] if d 1 mod 4 and is Z[ d 2 ] otherwise). These, and related rings, are the topic of algebraic number theory. Polynomial rings of course you know. If K is a field then any polynomial in one indeterminate X with coefficients in K can be factorised as a product of irreducible polynomials, and this decomposition is essentially unique. For instance if a, b,c K then ax 2 + bx + c either is irreducible or is a product of two (necessarily irredicible) linear factors (X α)(x β) where α,β are the roots of the quadratic polynomial (and these can be found using the quadratic formula). Did you ever wonder whether there could be two essentially different factorisations of such a polynomial? What about polynomials in more than one variable? - take a polynomial in, say, X, Y, Z with, say, rational coefficients: does it necessarily have a factorisation into irreducibles? (The answer is yes and that s easy to prove.); but is such a factorisation essentially unique? (The answer again is yes but this is not at all obvious and is quite hard to prove - but we ll do it.) 8

2 3.1 Unique Factorisation Domains Suppose throughout that R is a commutative domain. A non-zero, non-unit a R is irreducible if whenever we have a = bc then either b or c is invertible (so the other is, by definition of associate, an associate of a). It follows that a is irreducible iff whenever we have a factorisation a = bc, one of b, c is an associate of a (it follows that the other is a unit by (the argument of) 2.6). For instance, if K is a field then a non-constant (note that implies, non-zero, non-unit) polynomial f K[X] = K[X 1,...,X t ] is irreducible if f cannot be written as a product of two polynomials of smaller degree: f = gh implies g or h is a constant (i.e. an element of K). The, easy, proof of this uses the fact (8.9) that if f = gh then deg(f) = deg(g) + deg(h). Note that non-zero polynomials f and g have the same degree if they are associates. Examples Let K = R, f = X R[X]. Since the square of a real number is always positive, X has no real root, hence has no linear factor. Hence f is an irreducible element of R[X]. 2. Let K = C, f = X C[X]. Now we have X = (X i)(x + i) where i is a square root of -1. So f is not irreducible as an element of C[X]. 3. Let f = X 2 +Y 2 1 K[X, Y ] where K is any field of characteristic 0 (such as Q, R or C). Is f irreducible? Suppose not, say f = gh with neither g nor h a constant. Since deg(f) = 2, it must be that deg(g) = 1 = deg(h). Write each of g, h as a sum of its homogeneous parts: g = g 1 + g 0 and h = h 1 + h 0. Then we have X 2 +Y 2 1 = (g 1 +g 0 )(h 1 +h 0 ) = g 1 h 1 +(g 1 h 0 +g 0 h 1 )+g 0 h 0 where g 1 h 1 is homogeneous of degree 2, g 1 h 0 +g 0 h 1 is homogeneous of degree 1 and g 0 h 0 is homogeneous of degree 0. Recall, 8.1, that the decomposition of a polynomial into homogeneous components is unique so we obtain: X 2 + Y 2 = g 1 h 1,0 = g 1 h 0 + g 0 h 1, 1 = g 0 h 0. From the last equation we get h 0 0 so we can divide by h 0 in the second equation, re-arrange and obtain g 1 = ( g 0 /h 0 )h 1. Then substitute this into the first equation to get X 2 + Y 2 = ( g 0 /h 0 )h 2 1. Now, h 1 = ax + by for some a, b K and, if we square this and multiply by the scalar g 0 /h 0 we cannot get terms X 2 and Y 2 without getting a non-zero cross-term (scalar)xy (write this out carefully to see that, at this point, we re using that the characteristic of K is 0, at least, that it s not 2) - contradiction. So X 2 + Y 2 1 is irreducible (over any field of characteristic 2). Exercise 3.2. Show that X 2 + Y 2 1 Z 2 [X] is not irreducible. 4. In the ring Z[ 2] the element is irreducible. We show this by considering the norm of an element which, in this ring, is defined by N(a + b 2) = a 2 2b 2, so N( ) = 7. It is easily checked that the norm is multiplicative in the sense that N(rs) = N(r)N(s) for r, s Z[ 2], so if we have a factorisation = rs then N(r)N(s) = 7. Since 7 is prime (rather, irreducible) in Z, that implies that, say, r has norm ±1. If we show that N(r) = ±1 implies r is a unit then we will be done. Suppose r = a + b 2 with a, b Z. If N(r) = 1 then a 2 2b 2 = 1, so (a + b 2)(a b 2) = 1, so ra + b 2 is a unit. If N(r) = 1 then a 2 2b 2 = 1. In this case it s less obvious how to proceed and there are actually non-trivial solutions to the equation N(r) = 1: 9

3 e.g. N( ) = 1. But note that if N(r) = 1 then N(r 2 ) = (N(r)) 2 = 1 so, as seen just above, r 2 = ±1. Thus r 2 is a unit, hence (note, this is a general point) r is a unit, as required. We will show (3.24) that every polynomial in any number of variables has an essentially unique decomposition as a product of irreducible factors. A commutative domain R is said to be a unique factorisation domain if every non-zero, non-invertible element of R has an essentially unique factorisation as a product of irreducible elements. More formally: the commutative domain R is a unique factorisation domain, or UFD for short, if for every nonzero r R which is not invertible there are irreducible elements r 1,...,r k R such that r = r 1 r k (existence of irreducible factorisation) and, if also r = s 1 s l where the s j are irreducible elements of R, then k = l and there is a permutation σ of {1,...,k} such that, for each i, s σ(i) is associated to r i (uniqueness of irreducible factorisation). Example 3.3. Z is a unique factorisation domain. For instance take r = 24. Then 24 = ( 2) 2 ( 2) ( 3) is one irreducible factorisation. There are others, such as 24 = 3 ( 2) 2 2 but you can surely see how to permute the factors so that they match up as associated elements. Example 3.4. The ring Z[ 5], though a commutative domain, is not a UFD. For instance 21 = 3 7 = ( )(1 2 5) are, one may check, two factorisations into irreducibles but, again one may check, the irreducible factors do not match up into associated pairs. Another example in the same ring is 6 = 2.3 = (1 + 5)(1 5). To check that these really are different factorisations into irreducibles we have to show, say in the second example, that all of 2, 3, 1 + 5, 1 5 are irreducible and also that, for instance, 2 is an associate of neither nor 1 5. This is easy if we use the norm on Z[ 5]: this is the function N : Z[ 5] Z which is defined by N(a + b 5) = a 2 + 5b 2. It is easy to check that N(rs) = N(r)N(s) for all r, s Z[ 5]. Note also that if N(a+b 5) = a 2 +5b 2 = 1 then (a+b 5)(a b 5) = 1 so a+b 5 is invertible. Now, N(2) = 4 so if 2 factorises, say as 2 = rs with neither r nor s invertible, then N(r) = 2 = N(s) which is impossible (the equation a 2 +5b 2 = 2 clearly has no integer solutions). So we ve shown that 2 is irreducible. Turning to 1+ 5: this has norm 6 so a proper factorisation 1+ 5 = rs with neither of r, s invertible would be possible only if N(r) = 2 and N(s) = 3 or vice versa, but we ve seen already that N(r) = 2 is impossible. Therefore is irreducible and the proof for 1 5 is almost identical. Finally we have to check that 2 doesn t divide either 1 5 or 1 5: if it did then the norm of 2, which is 4, would divide the norm of or both of which equal 6 - contradiction, as required. (By the way, the general formula for the appropriate norm to use in a ring of the form Z[ d] where d Z is squarefree is N(a + b d) = a 2 db 2 ; you can check this has the required multiplicative property N(rs) = N(r)N(s). Recall that such a norm was used in the motivating example I gave at the start of the course: a semiproof for Fermat s assertion that the equation y = x 3 has for integer solutions only x = 3, y = ±5, and which involved moving to the ring Z[ 2]. That example can be found, with details sketched, on pp. 80/81 of Stewart and Tall; Algebraic Number Theory.) 10

4 Lemma 3.5. The ring Z of integers is a unique factorisation domain. If K is a field then K[X] is a unique factorisation domain. We have seen already (2.2, 2.7) that these are PIDs, so this will follow from We will show that if K is a field then K[X 1,...,X t ] is also a unique factorisation domain but the proof is considerably harder. Existence of a factorisation into irreducibles is not a problem - the proof is given next - it s the uniqueness of factorisation which is difficult. Theorem 3.6. Let f K[X]. Then f = f 1...f n where each f i is irreducible. Proof. If f is irreducible then we re already finished. Otherwise f = g 1 g 2 with deg(g 1 ),deg(g 2 ) < deg(f). By induction on the total degree of a polynomial (see 8.9), we may assume that each of g 1 and g 2 can be written as a product of irreducible polynomials. So, putting together these expressions for g 1, g 2 we get an expression for f as a product of irreducible polynomials. [Note how saying by induction on the degree shortcuts the keep splitting any reducible factors idea behind the proof.] The greatest common divisor of non-zero elements a, b R (a commutative domain) is an element c = gcd(a, b) R such that c a, c b and, for any d R with d a, d b we have d c. It is easy to check (exercise) that if a greatest common divisor of a and b exists then it is unique up to multiplication by an invertible element of R: that is, if c, d are both greatest common divisors of a and b then c d. Lemma 3.7. Let R be a unique factorisation domain. Then: (1) for all non-zero a, b R, gcd(a, b) exists and is unique up to association; (2) if a R is irreducible and if a bc then a b or a c. Proof. The general, and key, point is that if x y, say y = xz, then an irreducible factorisation of y is obtained by placing an irreducible factorisation of x next to an irreducible factorisation of z. In particular, if x y then each irreducible factor of x occurs, up to association, as an irreducible factor of y and the multiplicity of that factor (to association) in the factorisation of x is less than or equal to its multiplicity (up to association, but let s stop saying that) in any factorisation of y. (1) Factorise each of a and b and then let c be the product of all the factors, including multiplicity, that they have in common. Then proceed as you would if R = Z, using the key fact above to deduce that if d a and d a then d c. (2) Again, with the key fact, this is just as if you were dealing with ordinary integers. If, say, ad = bc, then write each of d, b, c as a product of irreducible elements; substitute back into the equation ad = bc; note that a occurs on the left-hand side hence, by unique factorisation, a occurs on the right-hand side; hence a is a factor of b or c, as required. 3.2 Prime ideals and prime elements Let R be a commutative ring. An element a R is prime if a is not invertible and whenever a bc (b, c R) then a b or a c. This is almost the usual definition in Z, except that in any domain 0 will count as a prime element. 11

5 An ideal P of R is prime if whenever a, b R with ab P then either a P or b P. It follows (by induction) that if P is a prime ideal and if some product a 1 a 2...a n lies in P then at least one of the a i lies in P. Examples 3.8. (1) Recall that in the ring Z of integers, every ideal is principal. By 3.9 below, n is a prime ideal iff n is a prime integer or n = 0. For instance 12 is not a prime ideal because we have yet neither 6 nor 4 is in 12. On the other hand, 17 is a prime ideal because if ab 17 then 17 ab and hence, since 17 is prime, 17 a or 17 b, that is, a 17 or b 17. (2) Let R = Q[X, Y, Z]. The ideal X is prime but XY is not. The ideal XY Z 2 looks as if it should be prime but you need to check that XY Z 2 is irreducible. Another prime ideal is X 2 + Y 2 (however this is no longer prime if regarded as an ideal in R = C[X,Y, Z] since in that ring we have the factorisation X 2 +Y 2 = (X+iY )(X iy )). The ideal XY +1, X 2 is not prime but the ideal X +1, Y +0.5, Z 2 is prime. What about XY Z 5, Y 3 X 5? It s not always obvious whether or not an ideal in a polynomial ring is prime; the Gröbner basis techniques that we will see later in the course provide algorithms to answer this kind of question. The next result illustrates that prime ideals generalise prime elements. The notion of an ideal arose exactly because there are rings of integers in number fields which don t have a good prime decomposition theory for elements but do have such a theory for ideals, with prime ideals (and note that a prime ideal is not required to be a principal ideal) replacing prime elements. Lemma 3.9. Let R be a commutative ring. Let p R, p 0; then p is a prime element iff the ideal, p, generated by p is a prime ideal. Proof. ( ) Suppose that p is a prime element and that ab p, so ab = pc for some c R. Since p is prime either p a, so a p, or p b, so b p, showing that p is a prime ideal. ( ) Suppose that p is a prime ideal and that p ab, so ab p. Then either a p, so p a, or b p, so p b, showing that p is a prime element. Lemma Let R be a commutative domain. Every non-zero prime element of R is irreducible. Proof. If p is prime and p = bc then either p b or p c, let s suppose the former. Thus we have both b p and p b so, by 2.1, b = p. Since R is a domain 2.6 applies and we deduce that p and b are associates. It follows (see the comment after the definition of irreducible) that p is irreducible. In Example 3.4 we can see that the converse of 3.10 is false: 2 is an irreducible element of Z[ 5] but 2 is not a prime(!) in that ring: 2 divides the product (1 + 5)(1 5) but, as shown in 3.4, divides neither of the factors. In UFDs, however, we do have the converse. Corollary Let R be a unique factorisation domain. Then every irreducible element is prime. That is, in a UFD, irreducible = prime (for non-zero elements). Proof. The first statement is exactly what 3.7(2) says. Then combine this with 3.10 above. 12

6 Next we show that every principal ideal domain is a unique factorisation domain. An ideal is said to be maximal if there is no ideal strictly between it and the whole ring; we consider these further in 4.3. Proposition Suppose that the commutative ring R is a PID and not a field. Then a principal ideal p of R is maximal iff the element p is prime and non-zero iff p is irreducible. Proof. Every maximal ideal is prime: a fact which we will prove later (4.17); so if p is maximal then, by 3.9, p is prime. For the converse, suppose that p is a prime element. Let r R \ p and let J = p, r denote the ideal generated by p and r together. It will be enough to show that J = R. Since R is a PID there is some s J such that J = s. Since p J we have p = st for some t R. Thus st p and, by 3.9 the ideal p is prime, so either s p or t p. Since we chose s outside p the first case is impossible, so we have t = up for some u R. But that gives us p = st = sup so (1 su)p = 0 and hence, since R is a domain, s is invertible. But that implies that J = p, s is the whole of R, as required. Every prime element is, by 3.10, irreducible so it remains to show that if p is irreducible then it is prime. By what has been shown already it will be enough to show that p is a maximal ideal. If not, then there is an ideal, by assumption principal, say s, with p < s < R. Since s R, s is not invertible. Since p s we have p = st for some t R. If t were invertible then p and s would be associates hence, by 2.6, would generate the same ideal, contradiction. So we have a proper factorisation of p, contrary to our assumption that p is irreducible. Corollary If R is a PID then a R is irreducible iff it is prime. Theorem Every PID is a UFD. Proof. Suppose that R is a PID. Existence of irreducible factorisations: suppose that a R is neither 0 nor invertible. If a is irreducible then we re done; if not, write a = bc where neither b nor c is an associate of a (nor invertible). By 2.1 we have a b and a c, with both inclusions being proper (by 2.6). If b is not irreducible then write it as b = b b with neither an associate of b, and similarly for c. Continue. If this splitting process stops at some point then we collect up all the irreducible factors and have an irreducible factorisation of a. If it doesn t stop then we obtain a strictly increasing sequence of principal ideals a b b... say. That contradicts the fact that every ideal of R is finitely generated - that is, R is a noetherian ring, since a noetherian ring can have no properly ascending sequence of ideals - a result, 4.1 which we will prove later. (If R is a Euclidean domain then we don t have to wait to finish this proof because, with notation as in the first paragraph and if ν is as in the definition of Euclidean domain, we have ν(a) > ν(b), ν(c) so, if the process did not stop we d get a strictly decreasing sequence of natural numbers, contradiction.) Uniqueness of irreducible factorisation: The proof is, now that we have that irreducible=prime (3.13), exactly as with integers: suppose that r = r 1 r k = s 1 s l where the r i and s j are irreducible elements of R. Since r 1 is irreducible it is prime so it divides one of the s j and hence, since s j is irreducible, r i and s j are associates. Cancel r 1 from each side of the equation 13

7 (if s j = ur 1 that will leave an extra factor u on the RHS but u is a unit, so can be absorbed into one of the other factors). Continue... In this way we pair up the factors r i with associate factors s k, as required for uniqueness. In particular, by 2.7, we deduce the following corollary. Corollary Every Euclidean domain is a UFD. If R is a UFD then the notions of greatest common divisor and least common multiple may be defined as usual: gcd(a, b) is computed by picking out the common factors in irreducible factorisations of a and b and, more generally, any finite set of elements of R has a gcd which may be calculated by using the (easily proved) fact that gcd(a 1,...,a n ) = gcd(...gcd(gcd(a 1, a 2 ), a 3 ),...,a n )...). Actually computing irreducible or prime factorisations may, however, be difficult in practice (this difficulty is the basis for a standard method of securely(?) encrypting electronic communications on the internet). But if R is a Euclidean domain it is not necessary to compute prime factorisations in order to compute greatest common divisors and least common multiples: the method based on the Euclidean algorithm works just as well in this generality as it does for the ring, Z, of integers (assuming that we are in a sufficiently explicit situation that computations are possible). The next result is another which generalises from the more familiar cases of integers and polynomial rings. Lemma Let R be a unique factorisation domain and let a, b R be irreducible and not associates. Then a b = ab. Proof. Since ab a and ab b certainly ab a b (that part is true for any commutative ring). For the converse, suppose that c a b : say c = c a, c = c b. By unique factorisation, it follows that each of a and b occurs (up to association) in a factorisation of c into irreducibles (which exists and is unique since R is a UFD). Since a and b are not associates of each other, these are different occurrences, so ab is a factor of c and hence c ab, as required. 3.3 The field of fractions of a commutative domain The process of forming fractions, by which Q is obtained from Z, may be generalised with any commutative domain R in place of Z. Let W = {(r, s) : r, s R,s 0} be the set of pairs of elements of R with the second element non-zero. The idea is that a pair (r, s) will represent the fraction r/s, i.e. rs 1 (which is why s has to be non-zero). But a given fraction has many different representations, e.g. 1/2 = 3/6 = 2/ 4, so we introduce an equivalence relation on W by defining (r, s) (t, u) iff ru = st (why? because r/s = t/u iff ru = st). It s easy to check that this is indeed an equivalence relation and we define Q to be the set of equivalence classes. Write (r, s)/ for the equivalence class of (r, s). Define an addition and multiplication on Q in the obvious way ( obvious when you think how it all works for ordinary fractions): define ((r, s)/ ) + ((t, u)/ ) = ((ru + st, su)/ ) and ((r, s)/ ) ((t, u)/ ) = ((rt,su)/ ). As usual, because we are defining operations on classes by reference to particular 14

8 representatives, we have to check that these operations are well-defined. This is easily done, then one checks that the result is indeed a ring structure on the set, Q, of equivalence classes, with (1,1)/ being the 1 and (0,1)/ being the 0. A bit tedious, but straightforward and left as an exercise. What about the connection with the original ring R? Well, the map ι : R S defined by r (r, 1)/ is easily checked to be an injective homomorphism. So we can think of R as sitting inside Q. The latter is called the field of fractions of R and also written Q(R). Another easy check shows that every element of Q has the form ι(r)ι(s) 1 where r, s R, so Q really is built up as fractions from R (at least, from the copy of R sitting inside it). Examples (1) Q(Z) = Q; (2) Q(K[X]) = K(X), the field of rational functions, if K is a field; (3) Q(Z[ 2]) = Q[ 2]; (4) Q(Z[i]) = Q[i]; (5) Q(Z[X]) = Q(X). In checking that these are correct one may make use of the fact (which you can try as an exercise) that if R is a domain and if θ : R L is any embedding of R into a field then there is a unique factorisation of θ : Q(R) L through the embedding ι : R Q(R) of R into its field of fractions. Our main example will be the case that R = K[X 1,...,X t ], the ring of polynomials in X 1,...,X t with coefficients in a field K, in which case Q(R) is the field of rational functions in X 1,...,X t with coefficients in K, usually denoted by K(X 1,...,X t ). The typical element of K(X 1,...,X t ) has the form p(x 1,...,X t) q(x where p, q K[X X 1,...,X t) 1,...,X t ] - for instance 2 Y +XZ XY Z Z 6 2 Q(X, Y, Z). 3.4 Gauss Lemma Suppose now that R is a UFD and that f R[X], f = a n X n + + a 1 X + a 0. Define the content, c(f), of f to be greatest common divisor of all its coefficients: c(f) = gcd(a n,...,a 1, a 0 ). This is defined only up to association. Clearly if f R[X] then f = c(f)f where c(f ) = 1. A polynomial f is said to be primitive if c(f) = 1 (more precisely, if c(f) 1). Example Take R = Z and let f = 15X 4 21x Then c(f) = gcd(15,21,6) = 3 (we can ignore 0 coefficients) and f = c(f)f = 3(5X 4 7X 2 + 2) where f = 5X 4 7X is a primitive polynomial. Lemma (Gauss Lemma) Let R be a unique factorisation domain and let f, g R[X]. Then c(fg) = c(f)c(g) (up to multiplication by an invertible element). In particular, any product of primitive polynomials is primitive. Proof. Let f = c(f)f, g = c(g)g : so f and g are primitive. Then we have fg = c(f)c(g)f g so, if we show that c(f g ) = 1 then we will have c(fg) = c(f)c(g). Therefore, what we have to show is that the product of two primitive polynomials is primitive. So assume now that f and g are primitive. We have fg = c(fg)h, say, and we want to show that c(fg) = 1. If not then, since R is a UFD, we can choose some irreducible factor p, say, of c(fg): so p divides every coefficient of fg. We will derive a contradiction from this. Let f = a n X n + + a 1 X + a 0 and g = b m X m + + b 1 X + b 0 with the a i, b j R. Then fg = n+m k=0 c kx k where c k = i+j=k a ib j. Choose i such that p divides a 0,...,a i but not a i+1 and choose j such that p divides b 0,...,b j but not b j+1. Because f and g are primitive we have i < n, 15

9 j < m. Consider the coefficient c i+j+2 = a i+j+2 b a i+2 b j + a i+1 b j+1 + a i b j+2 + +a 0 b j+i+2 of X i+j+2 in fg (notice, e.g., that i+j +2 could happen to be greater than n - any too-high-indexed coefficients we just take to be 0). Since p, by assumption, divides each coefficient of fg, p divides the left-hand side of this equation and hence divides the right-hand side. By choice of i, j we have that p divides every term of this sum except, possibly, a i+1 b j+1 so, since p divides the whole sum it must also divide a i+1 b j+1. But then, by 3.7(2), p divides either a i+1 or b j+1 - contrary to choice of i or j. This contradiction shows that c(fg) = 1, as required. 3.5 K[X] is a UFD If R is a commutative domain then we can form its field, Q = Q(R), of fractions, as in 3.3. Proposition Let R be a UFD and let Q be the field of fractions of R. Suppose that f R[X]. Then f is irreducible in R[X] iff f is irreducible when regarded as an element of Q[X]. Proof. Certainly if f has a proper factorisation in R[X] then it has a proper factorisation in Q[X]. For the converse suppose that f = gh is a factorisation of f in Q[X] with deg(g),deg(h) > 1. Write g = 1 d g, h = 1 e h where d, e R and g, h R[X] (for instance, take d to be the product (or the least common multiple) of all denominators of coefficients of g, similarly for h). Then write g = c(g )g, h = c(h )h. So we have g = c(g ) d g, h = c(h ) e h, c(g ) = 1 = c(h ). Note also that deg(g ) = deg(g) > 1, deg(h ) = deg(h) > 1. We have f = gh = c(g )c(h ) de g h and hence (de)f = c(g )c(h )g h. We want to divide now by de but we have to show that we can do this in R[X]. Taking the content of each side of this equation and using 3.19 we obtain (de)c(f) = c(g )c(h )c(g h ) = c(g )c(h )c(g )c(h ) = c(g )c(h ). Therefore we have c(g )c(h ) de = c(f) R and hence f = (c(f)g ) h is a proper factorisation of f in R[X], showing that f is reducible in R[X] if it is reducible in Q[X]. Corollary Let f Z[X] be a polynomial with integer coefficients. If f is irreducible in Z[X] then f is also irreducible in Q[X]. Proof. This is the special case of the previous result where R = Z. So if a polynomial with integer coefficients cannot be split over Z then it cannot be split over Q either. Example Let f = X X 2 3X 6 Q[X]. By the above result this is irreducible over Q iff it is irreducible over Z. If it were reducible over Z then, being a cubic, it must have a linear factor, say f = (X a)g for some integer a. Any such a must be an integer divisor of 6 and hence is ±1, ±2, ±3, ±6. If you substitute these values for X in f then in no case is the result equal to 0. We conclude that f has no integer root and hence is irreducible over Z and so, by the above result, is irreducible over Q. Theorem Suppose that R is a unique factorisation domain. Then R[X] is a unique factorisation domain. 16

10 Proof. Suppose that f R[X] is non-zero. If f is invertible (that is, a nonzero constant polynomial) then there is nothing to prove. So assume also that f is not invertible, hence deg(f) 1. Write f = c(f)f with f a primitive polynomial in R[X]. First we factorise c(f) and f separately and put their factorisations together to obtain a factorisation of f. Since R is a UFD there is a factorisation of c(f) as c(f) = c 1...c l with the c i irreducible elements of R. As you would expect, the c i are also irreducible elements of R[X] since if, say, c i = hh with h, h R[X] then, since 0 = deg(c i ) = deg(h)+deg(h ), we have that h and h are of degree 0 and hence are in R so, since c i is irreducible in R, one of h, h is invertible, as required. If f is already irreducible then fine; otherwise factorise it as f = gh with neither g, h invertible. That means that neither g nor h is a constant polynomial because 1 = c(f) = c(g)c(h). So deg(f) > deg(g), deg(h). Since, note, both g and h are primitive we can apply the same argument to each of them so, since degrees can t keep decreasing, eventually we obtain a factorisation f = f 1...f n of f into irreducible polynomials in R[X] each of degree at least 1. Putting these together we get a factorisation f = c 1...c l f 1...f n of f as a product of irreducible elements of R[X]. Now we have to prove uniqueness. So suppose also f = d 1...d k g 1...g m with the d i irreducible elements of R and the g j irreducible elements of R[X] of degree at least 1. Notice that if g R[X] is irreducible then c(g) = 1: otherwise we would have g = c(g)g with neither c(g) nor g invertible. So c(f i ) = 1 = c(g j ) for all i, j. It follows that d 1...d k = c(f) = c 1...c l and hence, since R is a UFD, l = k and the c i and d j are, up to rearrangement, pairwise associated. Therefore c 1...c l f 1...f n = f = d 1...d k g 1...g m = c 1...c l ug 1...g m for some invertible element u. Since R[X] is a domain we can cancel c 1...c l from each side to obtain f 1...f n = (ug 1 )g 2...g m. Thus we have two ways of writing f as a product of irreducible elements. Rewrite ug 1 as g 1 (invertible factors of R[X] can be ignored). Let Q be the field of fractions of R and notice that R[X] is a subset (even a subring) of Q[X]: so consider the two above factorisations of f as being in Q[X]. By 3.20 each f i and each g j is an irreducible element of Q[X]. Therefore f 1...f n and g 1...g m are two factorisations of f into irreducibles in Q[X]. Since Q is a field, Q[X] is a unique factorisation domain (3.5) and hence m = n and, after renumbering if necessary, we have f i = u i g i where u i is an invertible element of Q[X]. We can write each u i as hi a i with h i R[X] and a i R, and hence h i R (since any invertible element of Q[X] is a constant). So a i f i = h i g i and hence a i = c(a i f i ) c(h i g i ) = h i (c(g i ) = 1 since g i is irreducible in R[X]). Hence a i h i and so we can cancel a i to get f i = u i g i for some unit u i R. Therefore f i and g i are associates, which is what we wanted to show (that is, unique decomposition of f and hence of f). Corollary Let K be any field. Then K[X 1,...,X t ] is a unique factorisation domain. That is, if f K[X 1,...,X t ] is a non-constant polynomial and if f = af 1...f n = bg 1...g m where a, b K and each f i, g j is a monic irreducible polynomial then a = b, m = n and there is a permutation σ of {1,...,n} such that f i = g σ(i) for each i = 1,...,n. Proof. That K[X 1,...,X t ] is a UFD follows from and induction. The second statement follows directly. 17

11 That is, every polynomial f may be written as a scalar times a product of monic irreducible polynomials and, except for re-arranging the terms, there is just one way of writing f in this form. Of course, we could as well assume in 3.24 that K is any UFD and obtain the same result by the same argument. Example You can check that X 4 Y 4 X 2 Y 2 Z 2 +X 2 Y 2 Z X 2 Y 2 Z 3 +Z 2 is equal to (XY Z)(X 2 Y 2 + Z 1)(XY + Z) and, with a bit of work, you can check that these three factors are irreducible. By 3.24 there is no other essentially different way of factoring this polynomial as a product of irreducible polynomials. Let f K[X 1,...,X t ]. Define the zero-set, V (f), of f to be the set of all t-tuples of elements of K which, when sustituted into f, give 0: V (f) = {(a 1,...,a t ) K t : f(a 1,...,a t ) = 0}. Example Let K = R, f = X 4 + X 2 Y 2 X 2 R[X, Y ]. Notice that f = X.X.(X 2 + Y 2 1) = ggh, say. We saw in 3.1 that X 2 + Y 2 1 is irreducible, so this is the irreducible decomposition of f. Suppose that (a, b) V (f). Then f(a, b) = 0 so, g(a, b)g(a, b)h(a, b) = 0. Hence either g(a, b) = 0 or h(a, b) = 0 so either (a, b) V (g) or (a, b) V (h). The converse is also true: if (a, b) V (g) or (a, b) V (h) then f(a, b) = 0 so (a, b) V (f). The same argument shows that this is true in general: the zero-set of any polynomial is the union of the zero-sets of its irreducible factors. In this case, we have that V (X 4 +X 2 Y 2 X 2 ) = V (X) V (X 2 +Y 2 1) = the union of the Y -axis with the circle of radius 1, centred at the origin (0,0). 3.6 Irreducible polynomials in K[X] Most of what is in this section is review of things you have seen before. 1. Suppose f 0, f K[X]. Then V (f) is a finite set - any non-zero polynomial in just one indeterminate has only finitely many roots. 2. Suppose that f 0, f K[X] and let a K. Then f(a) = 0 iff (X a) is a factor of f. 3. Recall that if f R[X] is a polynomial with real roots then f factorises as a product of linear and quadratic polynomials. That is: Proposition If f R[X] is irreducible then deg(f) 2. Recall also that the proof of this uses the fact that we can extend R to a field (C) over which every polynomial splits as a product of linear polynomials. 4. There is a formula for finding the roots (and hence the linear factors) of any quadratic polynomial. This was known (at least special instances) to the Babylonians around 2000BC. At the end of the 1400 s Luca Pacioli (at Venice) judged that a solution to cubic equations was impossible by the methods known at the time, yet in 1545 Girolamo Cardano in Milan published his book Ars Magna which contained a procedure for finding the solution for any cubic (this procedure was also found earlier, at least in part, by Scipione del Ferro at Bologna and by Niccolo Tartaglia at Venice). 18

12 The same book contained a procedure for finding a solution to the general quartic (=degree 4) polynomial, due to Cardano s pupil Ludovico Ferrari. In all these cases, the roots of the polynomial could be expressed by a formula involving the ordinary arithmetic operations, addition, subtraction, multiplication and division, together with the extraction of roots (think, for instance, of the formula for the solution of a quadratic). The question then was whether there exists such a type of formula for the roots of a general polynomial of degree 5 (=a quintic) and similarly for all higher degrees. Gradually it came to be suspected that there is no such formula and this was proved in the early 1800s by Galois and Abel. In fact, Galois gave the exact conditions for a given polynomial to be solvable by radicals. This proof arose out of consideration of symmetries among the roots of a polynomial or, as we would now express it, of consideration of permutation groups acting on the roots of a polynomial. Indeed, this work is often regarded as the birth of group theory. For the history and a good deal of the mathematics, see Tignol s book. 5. Recall the fundamental theorem of algebra. Theorem Let f R[X] or C[X] be a non-zero polynomial of degree d. Then, regarded as an element of C[X], f splits as a product of d linear factors. In particular, a polynomial f of degree d has exactly d roots, where we count repeated roots with their multiplicity (=the number of times the corresponding linear factor appears in the irreducible decomposition of f). 3.7 Tests for irreducibility Proposition (Eisenstein s Criterion for Irreducibility) Let f Z[X], say f = a n X n + + a 1 X + a 0 with a n 0. Suppose that there is a prime p Z such that p does not divide a n, p divides every other coefficient a i and p 2 does not divide a 0. Then f is irreducible (in Z[X], and hence, 3.21, in Q[X]). Proof. Suppose, on the contrary, that f = gh with deg(g) > 0, deg(h) > 0, say g = b m X m + +b 1 X +b 0 and h = c k X k + +c 1 X +c 0 with b m 0, c k 0, m 1, k 1. Notice that b m c k X m+k is the leading term of gh = f and hence b m c k = a n. Since p 2 does not divide a 0 but p does divide a 0 exactly one of b 0, c 0 is divisible by p. Say p b 0 and p c 0. Let i be minimal such that p does not divide b i - since b m c k = a n is not divisible by p neither is b m, so there is such a value i m < n. Consider a i = b i c 0 + b i 1 c b 0 c i : on the right-hand side each of b i 1,...,b 0 is divisible by p and also, by assumption, a i is divisible by p. Hence the remaining term b i c 0 is divisible by p. But p does not divide c 0 hence p divides b i - contrary to choice of i. This contradiction shows that f is irreducible. Examples (1) Let f = 2X 9 25X X 3 15X 30 Z[X]. Taking p = 5 we have that p does not divide the leading coefficient, 2, of f, p divides every other coefficient of f and p 2 = 25 does not divide the constant term of f. Hence, by 3.29, f is irreducible in the ring Z[X] and hence, by 3.21, f also is irreducible as a member of the ring Q[X]. It is not, of course, irreducible as a member of R[X] or of C[X]. 19

13 (2) Let p be a prime integer and consider Φ(X) = X p 1 +X p 2 + +X+1 Q[X]. We claim that Φ(X) is an irreducible polynomial in Q[X]. Notice that Φ(X) = (X p 1)/(X 1). Substitute X+1 for X to get Φ(X+1) = (X+1)p 1 X = X p 1 + ( ) p p 1 X p ( ( p 2) X + p 1). The leading coefficient, 1, is not divisible by p, each binomial coefficient ( p i) is divisible by p and the constant coefficient, p, is not divisible by p 2. Hence, by Eisenstein s Criterion, 3.29, Φ(X + 1) is irreducible. But this means that Φ(X) is irreducible (because, if Φ(X) factorised then so would Φ(X + 1) - think about it). (3) Eisenstein s criterion is a test which can be used to show that some polynomials are irreducible: it cannot be used to show that a polynomial is reducible. For instance, let f = X 3 9 Q[X]. Eisenstein s criterion cannot be applied to this polynomial, nevertheless, it is irreducible because the (real) cube root of 9 is irrational. (We will give a proof of this in the example which follows the next result.) Lemma Suppose that f Z[X] is of degree d and suppose that n 2 is an integer such that the reduct of f modulo n is irreducible and has degree d. Then f is irreducible (in Z[X], hence, by 3.20, in Q[X]). Proof. Suppose, for a contradiction, that f = gh with deg(g) = k, deg(h) = l k, l < d = deg(f). Write [f] n for the reduct of f modulo n, that is, the polynomial in Z n [X] which is obtained by replacing each coefficient in f by its congruence class modulo n, and similarly for g, h. Then we have [f] n = [g] n [h] n. Clearly, deg([g] n ) k, deg([h] n ) l. By assumption, deg([f] n ) = d so, since deg([g] n )+deg([h] n ) = deg([f] n ) = d we must have deg([g] n ) = k, deg([h] n ) = l. But that means that [f] n is not irreducible, contrary to hypothesis. Hence f must be irreducible, as claimed. Notice that the condition that the reduct of f modulo n also has degree d is equivalent to the condition that n should not divide the leading coefficient of f. Examples (1) In Example 3.30(3) we claimed that the cube root of 9 is irrational. We will prove this now by showing directly that X 3 9 is irreducible as an element of Q[X]. So consider the reduction of this polynomial modulo the prime 7: you can check that each value 0,1,...,6 gives a non-zero value for this polynomial in Z 7 so X 3 9 has no root in Z 7 and hence no linear factor. Therefore, being of degree 3, the reduction of X 3 9 modulo 7 is irreducible as an element of Z 7 [X]. Therefore, by 3.31, X 3 9 Q[X] is irreducible. It follows that X 3 9 has no rational root and we conclude that the real cube root of 9 is, indeed, irrational. (2) Consider X 3 +X 2 +X 1. Consider the reduction of this modulo 3 and set X = 0,1,2 in turn. The values obtained in Z 3 for the polynomial are 2,2,1, none of which is 0. Hence, since the polynomial is of degree 3, it has no linear factor, hence is irreducible and hence X 3 + X 2 + X 1 Q[X] is irreducible. (3) Consider X 5 +X 2 +X 1. Consider the reduction of this modulo 3 and set X = 0,1,2 in turn. The values obtained in Z 3 for the polynomial are 2,2,1, none of which is 0. It follows that X 5 +X 2 +X 1 does not have any linear factor but we cannot conclude from this that the polynomial is irreducible: there is still the possibility that it factorises as gh where deg(g) = 3, deg(h) = 2. We can proceed as follows: we can take g, h to have the forms g = X 3 + ax 2 + bx + c, h = X 2 + dx + e. Multiply these together and compare coefficients in the 20

14 equation gh = f. We get the following system of equations: a+d = 0, ad+b = 0, ae + c + bd = 1, cd + be = 1, ce = 1. The first two equations allow use to get rid of d and b and the last one lets us write e in terms of c. This leaves two equations in a and c to be solved. But, since we can work in Z 3 we know that a = 0,1 or 1 so, trying these in turn we find, in each case, that no value of c will solve the equations (exercise: do these computations). We conclude that X 5 +X 2 +X +1 is, indeed, irreducible in Z 3 [X] and hence, by the result above, is irreducible in Q[X]. Example This example shows why, in 3.31, we need the hypothesis that deg([f] n ) = deg(f). Let f = 3X 3 X 2 6X + 2. Is f irreducible? If we reduce modulo 3, we have [f] 3 = X Now this is irreducible: [f] 3 (0) = 2 0, [f] 3 (1) = 1+2 0, [f] 3 (2) = so [f] 3 has no linear factor and hence (since it has degree just 2) is irreducible. We cannot, however, conclude from this that f is irreducible. In fact f = (3X 1)(X 2 2) = gh, say. Reducing modulo 3 we have [g] 3 = 1 - a scalar. So the reducibility of f is hidden when we take reducts modulo 3. 21

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