Show if these Gaussian integers are irreducible or not

Question

Gaussian integers are the set $\mathbb{Z}[i]$ such that $\mathbb{Z}[i] = \{ a + bi | a, b \in \mathbb{Z} \}$. Unique factorisation does hold over the Gaussian integers.

(a) Which of the following are irreducible in $\mathbb{Z}[i]$: $ 4 , 2, 1+i$ ? ( prove that the given element is irreducible or write it as product of two non units.)

(b) Express each of the following elements of $\mathbb{Z}[i]$ as a product of irreducibles: i. $6$ ii. $1+3i$

Right for (a) $4=2\cdot 2=(1+i)(1+i)$ hence not irreducible $2=(1+i)(1-i)$.

Suppose that $(a+bi)(c+di) = 1 + i $ for some integers $a,b,c,d$.

Taking conjugates yields $(a-bi)(c-di) = 1 - i.$

Multiply the equations together: $(a^2 + b^2)(c^2 + d^2) = 2 $

Now, we have an equation in the integers. Since $2$ is prime, we have $a^2 + b^2 = 1$ and $c^2 + d^2 = 2$ or vice versa.

The first equation implies that $a = \pm1$ and $b = 0$ or $a = 0$ and $b = \pm1$ $\Rightarrow a + bi = \pm1$ or $\pm i$, a unit. The second implies that $c= \pm1$ and $d= \pm1$ hence $c+di=(1-i)$ or $(1+i)$ which are units. So $1+ i$ must be irreducible.

Is that right?

(b) $6=2\cdot3$ from previous results, $ 2=(1+1)(1-i)$ so can be written as product of irreducibles. But what about $3$? How can I show it is irreducible?

$1+3i=(1+i)(2+i)$. $1+i$ is an irreducible, as previously shown. Is $2 +i$, too?

Answer 1

(a) yes

with regard to "3": Suppose $$(a+ib)(c+id) = 3$$ Then $$(a^2+b^2)(c^2+d^2)=(a+ib)(a-ib)(c+id)(c-id) = 9$$ Because the prime factorization of 9 over the integers is 3*3 it follows that $$a^2+b^2$$ is either 1, 3, 9.

Case 1: $$a^2+b^2=1$$ then $$a+ib$$ is a unit

Case 2: $$a^2+b^2=9$$ then $$c+id$$ is a unit

Case 3: $$a^2+b^2 = 3$$ with integers a and b. This is impossible.

Thus 3 must be irreducible.

Answer 2

There are a few facts that are useful here.

One is that if the norm of an element is a prime integer, then the element is irreducible in the Gaussian integers. This shows that $1 \pm i$ and $2 \pm i$ are irreducible. (But note that the converse does not hold; $3$ is irreducible in the Gaussian integers (see below), but has norm $9$.)

The other is that there no elements of norm $\equiv 3 \pmod{4}$; just check that the sum of two squares is congruent to $0, 1, 2$ modulo $4$. This shows that all prime integers which are congruent to $3$ modulo $4$ remain irreducible in the Gaussian integers.

Answer 3

Suppose $(a+bi)(c+di)=3$ with $a+bi$, $c+di$ not units. Take norms to get $(a^2+b^2)(c^2+d^2)=9$. It follows that $a^2+b^2=3$ and $c^2+d^2=3$, a contradiction.

Answer 4

Claim: $4$ is reducible, and $2$ and $1+i$ are irreducible.

To see why $4$ is reducible take norms and conclude neither of the factors if $4=ab$ are units.

To see why $2$ is irreducible suppose $2=ab$ then

$$N(2)=4 = N(a)N(b)$$

so either $N(a)=1$,$N(b)=4$ (WLOG) we are done as $a$ is a unit.

If theyre both $2$ then the norm of no element in your ring is 2, ill leave it to you to show that.

To see why $1+i$ is irreducible, it has the same norm $2$ but the norm of no element is $2$ thus $1+i$ is irreducible.