Exercises on euclidean domains

Exercise 1. Let a a non null element of an integral domain with unit. Prove that a is a prime element iff (a) is a prime ideal.

Exercise 2. Consider the equivalence relation of being associates: prove that the class containing an invertible element is made of all and the only invertible elements of the ring.

Exercise 3. Determine all invertible elements of ℤ[√-7] = {a + bi√7 | a,b ∈ ℤ}.

Exercise 4. Let R the ring of Gauss' integers. Let I = (13) and J = (3 − 2i). Which one is the maximal ideal?

Exercise 5. Prove that in ℤ[√-5] the ideal I = (3,√-5 − 1) is not a principal ideal. Deduce that ℤ[√-5] is not an euclidean domain.

Exercise 6. Among the elements ℤ[√-7], which of them are associates.

3 + 4√-7,   −3 + 4√-7,   −3 −4√-7

Exercise 6. In ℤ[i] find the gcd of 7 + 3i and 5 −8i, in case there exists one.

Exercise 7. In ℤ[i] determine the gcd of 4 + 4i and −5 +7i, in case there exists one.

Exercise 8. In ℤ[i], tell if 3 is a prime element.

Exercise 8. Study the quotient ℤ[i]/(2).

Solutions

  1. Let R the integral domain with 1 then we have (a) = {ar | r ∈ R}.

    • a prime element   ⇐⇒   [(a | bca| b or a|c)]

    • a prime ideal   ⇐⇒   [bc ∈ (a) ⇒ b ∈ (a) or c ∈ (a)]

    From the second condition we've that either c = ar or b = ar so either a| c ore a| b, hence a is prime.

  2. Let a an invertible element (unit). Associates elements to a are those elements b such that: b = au, with u invertible elements. We know that the product of two units is a unit, so b is as well invertible.

  3. If the element is invertible its valuation is 1 that is N(a + b√-7) = a2 + 7b2 = 1. So the only invertible elements are those with b = 0 and a = ±1, hence ±1.

  4. 13 = (3 − 2i) (3 + 2i), thus 13 is reducible and I = (13) is not maximail. 3 − 2i is irreducible thus (3 − 2i) is maximal and (13) ⊂ (3 − 2i).

  5. Principal ideal, means that it can be generated by a single element z in R. Supposing (z) exists, there is a z = a + b√(-5), such that 3 ∈ (z) and -1+√(-5) ∈ (z). This means that we could find

    3 = zz',   √-5 − 1 = zz'', with z',z'' ∈ ℤ[√-5].

    Taking the norm, we get two multiplicative identities in ℤ,

    9 = N(z) ⋅ N(z'),   6 = N(z) ⋅ N(z'')

    which imply N(z) | gcd(9,6) = 3, then N(z) must be 1 or 3. The equation N(z) = N(a + b√-5) = a2 + 5b2 = 3, has no solutions in ℤ, so N(z) = 1, which forces z = 1 or -1. But 1 ∉ I since it would imply that

    1 = 3(a + b √-5) + (√-5 -1) (c + d √-5)   with a,b,c,d in Z

    1 = 3a + 3b √-5 + √-5c -5d -c -d√-5

    1 = 3a -5d -c + √-5 [3b -d + c]

    equivalent to the system

    1 = 3a -5d -c

    0 = 3b -d + c

    1 = 3a + 3b - 6d. Clearly impossible: 3a + 3b - 6d is clearly a multiple of 3 which 1 is not.

    Another way to see that the system is impossible: if we analyze the system mod 3, we have an equivalent congruence system

    1 ≡ 3a -5d -c (mod 3)

    0 ≡ 3b -d + c (mod 3)

    now using the fact that 3 = 0 (mod 3)

    1 ≡ -5d -c (mod 3)

    0 ≡ -d +c (mod 3)

    d ≡ c mod 3. Hence 1 ≡ -6d mod 3 equal to 1 ≡ 0 mod 3 which is an absurd, so the system has no solutions. If one ideal is not principal it's not PID.

  6. We first find the units of R. Let a + b √-7 be a unit;

    (a + b √-7)(c + d √-7) = 1

    Taking the norm, we obtain

    (a2 + 7b2)(c2 + 7d2) = 1

    Since a,b,c,d are integers, the only possibility is b = d = 0, a2 = c2 = 1, so that a = ±1. So the units are 1 and −1, and the associates of an element z are z and −z. Thus the associates are 3 + 4√-7 and -3 -4√-7.

  7. We first divide one Gaussian integer by the other in ℂ, to give

    (5-8i)/(7+3i) = (5-8i)(7-3i)/(7+3i)(7-3i) = 11/58 -71/58 i

    The Gaussian integer "nearest" to this is -i, so we take the quotient to be -i. This gives the first division in ℤ[i] to be

    5−8i = −i(7+3i) + (2 − i)

    with 2 − i as remainder. Next we divide 7+3i by 2-i in ℂ to give

    (7+3i)/(2 -2i) = (7+3i)(2+i)/(2-i)(2+i) = 11/5 + 13/5 i

    The Gaussian integer nearest to this is 2+3i, which we take as the quotient when 7+3i is divided by 2 -i in ℤ[i], to give

    7+3i = (2+3i)(2-i) − i. This is the second division ℤ[i]. It's not necessary to carry out a further division, since the remainder in the second division is -i, a unit, and so will divide into 2-i with zero remainder. Hence the gcd of 7+3i and 5-8i is -i.

  8. Using the euclidean algorithm, we first divide −5 +7i by 4 + 4i

    (−5 +7i) / (4+4i) = (−5 +7i) (4 - 4i)/ (4+4i) (4-4i) = (-20 + +48i + 28) / (16 + 16) = (8+38i)/32 = 1/4 + i3/2

    The gaussian integer 'nearest' to this is i, so we take the quotient to be i; to find the remainder you do (-5+7i)-(4+4i)*i = −1+3i. Now you've reduced your original problem to gcd(4+4i,-1+3i).

    2nd iteration, (4+4i)/(-1+3i) = 4/5 - (8/5)i. pick 1-2i. Pick 1-i we have: (4+4i)-(-1+3i)*(1-i)=2. So the gcd us (1-i).

  9. If 3 is reducible then 3 = ab where N(a) > 1 and N(b) > 1. Since N(3) = 9

  10. They are [0], [1], [i], [1+i].

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