rings lect 4

8/6/2019 Rings Lect 4

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4 Divisibility in integral domains

We have until this moment used Z and K[X] both as rich sources of examples, and alsoas prototypes: we have tried to develop ring theory in a way that captures the essential

properties of these structures. In this section we will try to generalize the divisibility andfactorization properties of these rings in a more general settings. For example we would liketo generalize the Fundamental Theorem of Arithmetic and the Euclidean Algorithm fromthe case ofZ to more general rings. We begin with introducing the appropriate terminologyin the abstract setting of integral domains.

4.1 Divisibility

Definition 1. Let R be a ring, and let a, b R. We say that b divides a (and write b|a) iffor some c R we have thata = bc. We will also callb adivisor ofa in these circumstances.

So for example the units are the divisors of 1, and everything divides 0.

Definition 2. LetR be a ring, and let a R. An element a of R is called an associate ofa if for some unit u R we have that a = ua.

We remark that this is an equivalence relation.For example the associates of n Z are n. Note that unique factorization in Z is not

really unique, for example 6 = 2(3) = (2)3. Of course this non uniqueness up to a signis quite harmless. In the unique factorization theorem that we will prove later uniquenesswill fail up to units rather than signs, which is still not bad. As we want to think of 2 and2 as the same factor of6 associate elements are the same as far as divisibility goes forgeneral rings.

Remark 4.1.1. If R is an integral domain and a, b R are such that a|b and b|a then a, bare associates.

Indeed if both a, b are 0 they are clearly associates. Otherwise lets say that b = 0. Sincea|b and b|a we have b = au and a = bv for some u, v R. Substituting a in the first equalitywe get

b = (bv)u b b(vu) = 0 b(1 vu) = 0 1 vu = 0So u, v are units and a, b associates.

Definition 3. Let R be a ring, and let a R, a = 0. We say that a is irreducible in R ifa / R and

a = xy = x is a unit, or y is a unit.Definition 4. Let R be a ring, and let a R \ {0} but a R . We say that a is prime inR if

a|xy = a|x or a|y.Essentially, irreducible elements cant be factorised further; prime elements are those

which only divide products of which they already divide one factor.In general we have this:

Proposition 4.1.2 (Prime Irreducible). Let R be an integral domain, and let x R beprime. Then x is irreducible.

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Proof. Suppose that x = yz. Then x = x1 so that x|x = yz. By the definition of primewe will get x|y or x|z; suppose the former, that y = xt say. Then x = yz = xtz, and sox(1tz) = 0. As x = 0 we get tz = 1, and z is a unit. Hence x is by definition irreducible.Definition 5. Let R be a ring, and let a, b

R. We say that d is a highest common

factor of a, b if

(i) d|a and d|b;(ii) e|a and e|b implies that e|d.

So for example in Z a highest common factor of 9 and 6 is (3); another is 3. Note thatin general there is no guarantee that highest common factors exist.

Proposition 4.1.3. LetR be an integral domain and let a, b R \{0}. Suppose that d1 andd2 are highest common factors of a, b. Then d1 and d2 are associates.

Proof. It is clear that neither d1 nor d2 is zero, as a and b are multiples of each.By condition (i) applied to d1 we see that d1|a and d1|b. So apply condition (ii) to d2using e = d1 to get that d1|d2. Similarly we get d2|d1. So by remark 4.1.1 we have that d1, d2are associates.

Proposition 4.1.4. Let R be an integral domain and let a, b R \ {0}. Suppose that d1is a highest common factor of a, b and that d2 is an associate of d1. Thend2 is a highestcommon factor too.

Proof. Suppose that d2 = d1u for some u such that uv = 1.For condition (i) note that a = d1x1 and b = d1y1, so that a = d2(vx1) and b = d2(vy1).For condition (ii) suppose that e|a and e|b. Then we have that e|d1, or d1 = ez. Then

d2 = d1u = ezu and we are done.

In some rings there is a sensible way to choose a particular highest common factorin Zwe usually choose the non-negative associate, in K[X] the monic associateand call it thehighest common factor. But often we can work quite comfortably with the uncertainty ofup to a unit multiple.

4.2 Euclidean Rings

In order to generalize the theorems of Arithmetic to rings we have to restrict to a specialclass of rings: Euclidean rings.

4.3 Definition

What we must do is express abstractly what is going on in the Division Algorithm. Wedivide a by b = 0 and get a quotient q, leaving a remainder r which is smaller in someway (size, degree, . . . ) than the divisor b.

Definition 6. We say that R is a Euclidean Ring with Euclidean function d if

(a) R is an integral domain;

(b) the function d : R \ {0} N \ {0} satisfies:

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(i) for all x, y R \ {0} we have d(xy) d(y);(ii) for all a, b R with b = 0 there exist q, r R such that a = bq + r and r = 0 or

d(r) < d(b).

The element q = q(a, b) is called the quotient of a by b, and the element r = r(a, b) is calledthe remainder.

In our customary rather slovenly way we will say R is a Euclidean ring when the functiond is so obvious as to be understood without mention. However, see the last example belowfor a warning.

4.4 Examples

4.4.1 The Integers

The ring of integers, equipped with the function n

|n

|on the non-zero elements, is a

Euclidean Ring: we have known this since we learned about division, and we proved that itis true in the first year course1.

4.4.2 Rational Polynomials

The ring Q[X], equipped with the degree function f deg f, is a Euclidean Ring: weveknown this since we learned about long division, and we proved that it is true in the firstyear course2.

Of course the same thing is true for polynomials over any field.

4.4.3 The Gaussian Integers

Let Z[i] := {a + bi | a, b Z}. We call these the Gaussian integers. This subset ofC is clearlya subring ofC and as C has no zero-divisors it is actually an integral domain.

How can we measure size, and find a Euclidean function d? The obvious choice is

d(a + bi) := |a + bi|2,

but does it satisfy the requirements?Condition (i) is easy:

d() = ||2 = ||2||2 = ||2d()

and as || = a2

+ b

2

for integral a, b we get ||2

1.Condition (ii) is more complicated; the argument is important as it can be used for certainother Z[

n] and not just Z[

1].So let := a + bi and := c + di = 0 be in the ring. Then in Q[i] := {x + yi | x, y Q}

we can rationalise the denominator and get that

=

a + bi

c + di=

ac + bd

c2 + d2+

bc adc2 + d2

i =: x + yi

say.

1On the course website there will be a proof.2On the course website there will be a proof.

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This is the exact quotient in Q[i], but what is the best we can do in Z[i]? The nearestwe can get is the number := m + ni, where m is the integer nearest x, and n the integernearest y; note that |x m| 1

2and |y n| 1

2.

Now

= x + yi = m + ni + (x m) + (y n)iand multiplying by we get

= (m + ni)+ ((x m) + (y n)i) .Put as quotient q(, ) := m + ni Z[i]. As remainder we then would have

r(, ) := ((x m) + (y n)i) = q(, ) Z[i].We now compute

d(((x m) + (y n)i) ) = | ((x m) + (y n)i) |2

= | ((x m) + (y n)i) |2

||2

= ((x m)2 + (y n)2) ||2 1

2||2

< d()

and see that condition (ii) is satisfied.

4.4.4 Fields

Let K be a field, and define d : K \ {0} N \ {0} by d(x) = 1. It is then trivial to see thatwe have a (very dull) Euclidean Ring where all the remainders are 0.

4.4.5 The Integers, but not as we know them

The ring of integers, equipped with the function

d(n) = the number of digits when |n| is expressed in base 2,is a Euclidean Ring.

For condition (i), note that

2M + lower powers of 2 2N + lower powers of 2 = 2M+N + lower powers of 2

and so d(xy) = d(x) + d(y) d(y).

For condition (ii), all is clear if a = 0, or indeed if d(a) < d(b); just take q = 0 andr = a. So argue by induction on d(a). Suppose that a =

2M + lower powers of 2

, and

b =

2N + lower powers of 2

with = 1 and = 1. We are assuming M N, soconsider a =

a 2MNb + 2MNb. The number a 2MNb requires fewer than

M = d(a) binary digits, so we can find q and r such that

a 2MNb = bq + rand r = 0 or d(r) < d(b).

Taking q = q + 2MN (=q 2MN) gives what we need.

This example shows that we need to take care when we say R is a Euclidean Ring.

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4.5 Units and Associates

If we are interested in factorisations the first thing we need to deal with are the factorisationsof the identity: we must find the units of the ring. So let R equipped with d be a Euclideanring.

Lemma 4.5.1. For all a R, d(a) d(1).Proof. Condition (i) applied to a = a 1 gives this at once.Lemma 4.5.2. For all units u R, d(u) d(1).Proof. Condition (i) applied to 1 = v u gives this at once.Lemma 4.5.3. For all x R such that d(x) = d(1), we have that x is a unit.Proof. Use condition (ii) to get q and r such that

1 = xq + r with r = 0, or d(r) < d(x).

If r = 0 then we have by hypothesis d(r) < d(x) = d(1); this contradicts Lemma 1. So weget exact division, 1 = xq and x is a unit.

To summarise:

Proposition 4.5.4 (Units of a Euclidean ring). Let R equipped with d be a Euclidean ring;then the group of units is given by

R = {x R | d(x) = d(1)} .

In fact we may arrange things so that we have d(1) = 1. For suppose that d(1) = k + 1for k N. Weve just seen that for all a R we have that d(a) d(1), so the functiond : R \ {0} N \ {0} by d : a d(a) k is well-defined. It is clear that it also satisfiescondition (i) and condition (ii) with the same quotient and remainder.

The following is also useful about a Euclidean ring R:

Lemma 4.5.5. Let u R and a R. Then d(ua) = d(a).Proof. Let v a be such that vu = 1. We then have by condition (i) that d(a) = d(vua) d(ua) d(a); equalities rule, and d(a) = d(ua).

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The following subsections, establishing that the integers and polynomial rings over fieldsare Euclidean are included purely for completeness; they were covered in the first year course.

Revision: integers

Let d : Z \ {0} N \ {0} be given by d(n) = |n|.For condition (i) we can use the properties of and |n| we developed in Analysis I and

see that d(xy) = |xy| = |x||y| = |x|d(y) d(y).For condition (ii) note that

a = bq + r (a) = b(q) + (r)

and |r| = | r|; so it is enough to deal with the case a 0.We can argue by (strong) induction on a; the result is true for a = 0 if we take q = 0 and

r = 0. Indeed the result is true for a < |b|, just take q = 0 and r = a. For a |b| note thata = (a |b|) + |b|, so we can use the inductive hypothesis to get (a |b|) = bq + r with r = 0or |r| < |b|. Taking q = q 1 as case may be completes the proof.

Revision: polynomials

Let d : K[X] \ {0} N \ {0} be given by d(f) = deg(f), the degree of f.For condition (i) note that if f =

mk=0 akX

k and g =n

k=0 bkXk, with am = 0 and

bn = 0 then fg =m+n

k=0 ckXk where cm+n = ambn. Hence deg f g = deg f+deg g deg g.

For condition (ii), again it is clearly true if f = 0; just take q = r = 0. Otherwise notethat if the top coefficient of f is = 0 and the top coefficient of g is = 0, then

f = gq + r 1

f =1

g

q

+1

r

;

clearly deg1

f

= deg f, deg1

g

= deg g and deg1

r

= deg r. Hence we may assume

that f, g are monic, that is have top coefficients 1.Now we may argue by induction on deg f; if deg f = 0 or more generally deg f < deg g we

put q = 0 and r = f and are done. Otherwise, note that f(X) =

f(X) Xdeg fdeg gg(x)+Xdeg fdeg gg(x), and that deg

f(X) Xdeg fdeg gg(x) < deg f. By the inductive hypoth-

esis we can then find q and r such that

f(X) Xdeg fdeg gg(X) = g(X)q(X) + r(X)

with deg r < deg g or r = 0. Now put q(X) = q(X) + Xdeg fdeg g and we are done.

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Which of the following are true?

1. IfK is a field then any non zero element ofK divides any other non zero element ofK.

2. Ifa, b are associates then a|b and b|a.3. Ifa is irreducible and a, b are associates then b is also irreducible.

4. Ifa is prime and a, b are associates then b is also prime.

5. a|b if and only if (a) (b).6. The ideal < X > + < Y > ofR[X, Y] is principal.

7. R[X, Y] is a Euclidean domain.

8. IfR is Euclidean ring with Euclidean function d then d(a) > d(1) for all a = 0, a R.

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rings lect 4

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