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ALGEBRA HOMEWORK 6

NATHAN ST. JOHN

(1) Two nonzero ideals I and J are called coprime if IJ = I J.

a.) Prove that comaximal ideals are coprime.

Let I and J be comaximal ideals of D. We have IJ I J since idealsare closed under multiplication by elements of D.

For the other containment, let a

I

J. Choose i

I and j

J suchthat i+j = 1. Then ai+aj = a. Both ai and aj are in IJ because a IJ,so ai + aj IJ since IJ is an ideal. Thus, a IJ, completing the proof.

b.) Give an example to show that the converse is not true.

Consider the ring Z[x, y] and the ideals I = (x) and J = (y). Both IJand I J consist of all polynomials with terms of the form nxiyj (n Zand i,j > 0), so they are equal. Hence, I and J are coprime. They are notcomaximal, however, since 1 / (x) + (y).

c.) Give, with proof, a sufficient condition for the converse to be

true.

Suppose D is a PID. Let I = (a) and J = (b) be coprime ideals. ClearlyIJ = (ab). We claim that I J = (lcm(a, b)).

Let m = lcm(a, b). By definition, a | m and b | m, so m IJ. It followsthat (m) = I J.

Now let c I J. Since c I, a | c. Since c J, b | c. By the definitionof lcm(a, b), m | c, so c (m). We conclude that I J = (lcm(a, b)).

Now, since (m) = (ab), we have m = ab = lcm(a, b) gcd(a, b). Thisimplies that gcd(a, b) = 1, so there exist c, d D such that ac + bd = 1.Therefore, (a) + (b) = D, as desired.

(2) Prove that if gcd(a, b) exists for all a, b D, then lcm(a, b) existsfor all a, b D.

Let d = gcd(a, b). We claim that m = ab/d is lcm(a, b). Since d | a (andd | b, but we dont need that here), d | ab, so our proposed m is indeed anelement of D.

Date: November 19, 2009.

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2 NATHAN ST. JOHN

If we rewrite m as a(b/d) and b(a/d), we see that a | m and b | m, respec-tively. To verify the other condition on least common multiples, suppose

there is an n D such that a | n and b | n. We need to show that m | n.Consider the following series of equalities:

n = n dd ab

ab=

nd

ab ab

d=

nd

ab m.

If we can show that ab | nd, we will be complete, since this implies thatm | n. To this end, notice that gcd(na, nb) = n gcd(a, b) = nd. Also a | nand b | n, so we have ab | nb and ab | na. From the definition of greatestcommon divisor, ab | nd, as desired.

(3) If D is a UFD and I = (a), show that

J =

x

D : there exists a k > 0 such that xk

(a)

is principal.

Suppose a = se11 senn is an expression of a as a product of irreducibles

si such that si = sj unless i = j. We claim that (s1 sn) = J. Clearly(s1 sn) J since (rs1 sn)e1en (a) for any r D.

Now suppose x J and xk = tf11 tfmm is an expression of xk as a

product of distinct irreducibles. Since xk = ra for some r D, we havetf1

1 tfmm = rse11 senn . By uniqueness, we know that each si must appear

on the left hand side as an rj. Thus, the product s1 sn can be factoredout of the left. This shows that xk (s1 sn), as desired.

(4) Let D be a euclidean domain with valuation v and a and bnonzero elements of D. Show that v(a) = v(ab) if and only if b is aunit.

For the forward direction, suppose v(a) = v(ab) and choose q and r suchthat a = abq + r, where r = 0 or v(r) < v(ab). Ifr = 0, we are done sincethen a = abq, which implies 1 = bq. Now suppose v(r) < v(ab)). Noticethat a(1 bq) = r, so

v(a) v(a(1 bq)) = v(r) < v(ab) = v(a),a contradiction. Therefore, b is a unit.

For the reverse direction, suppose b is a unit. Then

v(a) v(ab) v(abb1) = v(a),so v(a) = v(ab).

(5) Use the domain of Gaussian integers to show that the remain-der given by the Euclidean algorithm is not unique.

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ALGEBRA HOMEWORK 6 3

Consider the following:

7 + 4i = (2

i)(3 + 2i) + (

1 + 3i)

7 + 4i = 2(3 + 2i) + 1

(6) Show that D =

a + b2 : a, b Z is a euclidean domain.

Let N(a + b2) = a2 + 2b2 and , D. We need to find q and r such

that = q + r and N(r) < N(s).Suppose = a + b

2 and = c + d2. Then

=

a + b2

c + d2 =

ac + 2bd

c2 + 2d2 e

+bc ad

c2 + 2d2 f

2.

Let e and

f be the closest integers to e and f, respectively. Define =(e e) + (f f)2 and = = (e + f2). It follows that = (e + f

2) + .All that remains is to show N() < N(). To this end, we show that

N(xy) = N(x)N(y), where x = x1 + x22 and y = y1 + y2

2. Considerthe following:

N(xy) = N

x1y1 2x2y2 + (x1y2 + x2y1)2

= (x1y1 2x2y2)2 + 2(x1y2 + x2y1)2= (x21 + 2x

22)(y

21 + 2y

22)

= N(x)N(y).

With this in hand, note that N() = N()N(). Now, N() < 3/4since |e e| and |f f| are both less than 1/2. Therefore, N() < N(),completing the proof.