# Irreducibility

a. Show that 1 + i is not a unit in Z[i]

b. Show that 2 is not irreducible in Z[i]

c. Show that 3 is irreducible in Z[i]

https://brainmass.com/math/linear-algebra/irreducibility-397853

## SOLUTION This solution is **FREE** courtesy of BrainMass!

Proof:

(a) Suppose 1 + i is a unit in Z[i], then we can find some a + bi in Z[i], such that

(1 + i)(a + bi) = (a - b) + (a + b)i = 1

Then we get

a - b = 1 and a + b = 0

We solve the system of equations and find that a = 1/2, b = -1/2

So a + bi = (1/2) - (1/2)i is not in Z[i], we get a contradiction.

Therefore, 1 + i is not a unit in Z[i].

(b) We know that 2 = (1 + i)(1 - i), both 1 + i and 1 - i are in Z[i]. Thus 2 is not irreducible in Z[i]

(c) Suppose 3 is irreducible in Z[i], then we can find two elements a+bi and c+di in Z[i], such that

(a+bi)(c+di) = (ac-bd) + (bc+ad)i = 3

Then we get ac - bd = 3 and bc + ad = 0. So bc = -ad. we have the following cases.

Case 1: a = 0. Then bc = 0, but b can not be zero, then c = 0. So ac - bd = -bd = 3

Then b = 3, d = -1 or b = 1, d = -3. So 3 = (3i) * (-i) = i * (-3i). But both i and -i are units, then

3 is irreducible.

Case 2: a is not 0, then c is not 0 either. Then bc = -ad implies b/a = -d/c. We also have

ac - bd = 3, then we 1 - (bd)/(ac) = 3/(ac), then -(bd)/(ac) = (b/a)^2 = 3/(ac) - 1 >=0

Then ac <= 3. If ac = 3, then (b/a)^2 = 0 and b =0, then d = 0. Thus 3 = ac. But 3 is irreducible

in Z, then 3 is also irreducible in Z[i]. If ac<3, no matter ac = 2 or ac = 1, 3/(ac) - 1 can not be

a square number and we get a contradiction.

Therefore, 3 is irreducible in Z[i].

https://brainmass.com/math/linear-algebra/irreducibility-397853