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In this question, we work with ?[i] = {a + ib | a, b ? ?} ? ?, i.e. the set of complex numbers with real and imaginary parts that are integers. Similarly to the previous problem set, we define the norm of z = a + ib to be N(a + ib) = a^2 + b^2 = zz?. As in last week's problem set, we have that N(z)N(w) = N(zw), and moreover the norm of a nonzero Gaussian integer is a positive integer. 1. Show that the units in ?[i] are those elements that have norm 1 (i.e. 1, -1, i, -i). Now suppose that z = a + ib is a nonzero Gaussian integer, and w is any Gaussian integer. We would like to show that there are Gaussian integers q, r such that w = qz + r with N(r) < N(z) (analogous to the division of integers, or polynomials over a field). The ratio w/z (which is supposed to approximate q) may not be a Gaussian integer, but it's a perfectly good complex number. Let us write w/z = r + is (where r, s ? ?). Let us choose r0 to be the integer nearest r (so that |r0 - r| ? 1/2), and let s0 be the nearest integer to s (so that |s0 - s| ? 1/2). 2. Show that N(w - (r0 + is0)z) < N(z). (Hint: this inequality doesn't use the fact that any of the quantities involved are Gaussian integers; you can multiply by N(1/z), which should simplify the problem considerably.) Thus we can take q = r0 + is0, and r = w - qz. Note that in the case of integers and polynomials over a field, we had that q and r are unique. In this case, we don't have uniqueness. For example, according to our criteria, when z = 2 and w = 1, we have w = 0 × z + 1, w = 1 × z + (-1). This means that (q, r) could be taken to be (0, 1) or (1, -1). Nevertheless, it's good enough for our purposes. 3. Show that every ideal of the Gaussian integers is principal. (Hint: mimic the proof for usual integers.) Since any subring of a field has no zero divisors, and 1 is a Gaussian integer, it follows that the Gaussian integers are a principal ideal domain. This also implies that the ring of Gaussian integers is a unique factorisation domain. Now we shift our focus to understanding the primes in the Gaussian integers. We will think of ? as the set of Gaussian integers with imaginary part equal to zero, and call such elements "ordinary integers" (and an "ordinary prime" is prime as an element of ?). 4. Suppose that z is a prime element of ?[i]. Show that z divides an ordinary prime number. (Hint: first find an ordinary integer divisible by z, and then consider the factorisation of that ordinary integer into ordinary primes.)

          In this question, we work with ?[i] = {a + ib | a, b ? ?} ? ?, i.e. the set of complex numbers with real and imaginary parts that are integers. Similarly to the previous problem set, we define the norm of z = a + ib to be

N(a + ib) = a^2 + b^2 = zz?.

As in last week's problem set, we have that N(z)N(w) = N(zw), and moreover the norm of a nonzero Gaussian integer is a positive integer.

1. Show that the units in ?[i] are those elements that have norm 1 (i.e. 1, -1, i, -i).

Now suppose that z = a + ib is a nonzero Gaussian integer, and w is any Gaussian integer. We would like to show that there are Gaussian integers q, r such that w = qz + r with N(r) < N(z) (analogous to the division of integers, or polynomials over a field).

The ratio w/z (which is supposed to approximate q) may not be a Gaussian integer, but it's a perfectly good complex number. Let us write w/z = r + is (where r, s ? ?). Let us choose r0 to be the integer nearest r (so that |r0 - r| ? 1/2), and let s0 be the nearest integer to s (so that |s0 - s| ? 1/2).

2. Show that
N(w - (r0 + is0)z) < N(z).
(Hint: this inequality doesn't use the fact that any of the quantities involved are Gaussian integers; you can multiply by N(1/z), which should simplify the problem considerably.)

Thus we can take q = r0 + is0, and r = w - qz. Note that in the case of integers and polynomials over a field, we had that q and r are unique. In this case, we don't have uniqueness. For example, according to our criteria, when z = 2 and w = 1, we have
w = 0 × z + 1, w = 1 × z + (-1).
This means that (q, r) could be taken to be (0, 1) or (1, -1). Nevertheless, it's good enough for our purposes.

3. Show that every ideal of the Gaussian integers is principal. (Hint: mimic the proof for usual integers.)

Since any subring of a field has no zero divisors, and 1 is a Gaussian integer, it follows that the Gaussian integers are a principal ideal domain. This also implies that the ring of Gaussian integers is a unique factorisation domain. Now we shift our focus to understanding the primes in the Gaussian integers. We will think of ? as the set of Gaussian integers with imaginary part equal to zero, and call such elements "ordinary integers" (and an "ordinary prime" is prime as an element of ?).

4. Suppose that z is a prime element of ?[i]. Show that z divides an ordinary prime number. (Hint: first find an ordinary integer divisible by z, and then consider the factorisation of that ordinary integer into ordinary primes.)

In this question, we work with ?[i] = a + ib | a, b ? ? ? ?, i.e. the set of complex numbers with real and imaginary parts that are integers. Similarly to the previous problem set, we define the norm of z = a + ib to be

N(a + ib) = a^2 + b^2 = zz?.

As in last week's problem set, we have that N(z)N(w) = N(zw), and moreover the norm of a nonzero Gaussian integer is a positive integer.

1. Show that the units in ?[i] are those elements that have norm 1 (i.e. 1, -1, i, -i).

Now suppose that z = a + ib is a nonzero Gaussian integer, and w is any Gaussian integer. We would like to show that there are Gaussian integers q, r such that w = qz + r with N(r) < N(z) (analogous to the division of integers, or polynomials over a field).

The ratio w/z (which is supposed to approximate q) may not be a Gaussian integer, but it's a perfectly good complex number. Let us write w/z = r + is (where r, s ? ?). Let us choose r0 to be the integer nearest r (so that |r0 - r| ? 1/2), and let s0 be the nearest integer to s (so that |s0 - s| ? 1/2).

2. Show that
N(w - (r0 + is0)z) < N(z).
(Hint: this inequality doesn't use the fact that any of the quantities involved are Gaussian integers; you can multiply by N(1/z), which should simplify the problem considerably.)

Thus we can take q = r0 + is0, and r = w - qz. Note that in the case of integers and polynomials over a field, we had that q and r are unique. In this case, we don't have uniqueness. For example, according to our criteria, when z = 2 and w = 1, we have
w = 0 × z + 1, w = 1 × z + (-1).
This means that (q, r) could be taken to be (0, 1) or (1, -1). Nevertheless, it's good enough for our purposes.

3. Show that every ideal of the Gaussian integers is principal. (Hint: mimic the proof for usual integers.)

Since any subring of a field has no zero divisors, and 1 is a Gaussian integer, it follows that the Gaussian integers are a principal ideal domain. This also implies that the ring of Gaussian integers is a unique factorisation domain. Now we shift our focus to understanding the primes in the Gaussian integers. We will think of ? as the set of Gaussian integers with imaginary part equal to zero, and call such elements "ordinary integers" (and an "ordinary prime" is prime as an element of ?).

4. Suppose that z is a prime element of ?[i]. Show that z divides an ordinary prime number. (Hint: first find an ordinary integer divisible by z, and then consider the factorisation of that ordinary integer into ordinary primes.)

Added by Josefina N.

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