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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*2=(1+i)(1+i) $hence not irreducible $2=(1+i)(1-i)$ 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?

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3 Answers 3

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.

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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.

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(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.

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