Question

Factor each of the elements below as a product of irreducibles in $\mathbf{\mathbb{Z}}\left[i\right]$,

1. 3
2. 7
3. $\mathbf{4}\mathbf{+}\mathbf{3}\mathit{i}$
4. $\mathbf{11}\mathbf{+}\mathbf{7}\mathit{i}$

Short answer

1. 3
2. 7
3. $(2-i)(1+2i)$
4. $(1+i)(1+2i)(1-4i)$

Step-by-step solution

(a)Step 1: Factor 3 as product of irreducibles in  ℤ[i]

Let $3=ab$then $N\left(a\right)N\left(b\right)=N\left(ab\right)=N\left(3\right)=9$

So $N\left(a\right)=13$or 9.

But $N\left(a\right)=3$is impossible as the equation $s^2+t^2=3$has no integral solution.

So, either role="math" localid="1654713332281" $N\left(a\right)$or $N\left(b\right)=1$.

Thus 3 is irreducible.

Therefore, factorization of $3=3$.

(b)Step 2: Factor 7 as product of irreducibles in ℤ[i]

Let $7=ab$ then $N\left(a\right)N\left(b\right)=N\left(ab\right)=N\left(7\right)=49$

So role="math" localid="1654713492487" $N\left(a\right)=1,7$or 49.

But$N\left(a\right)=7$ is impossible as the equation$s^2+t^2=7$ has no integral solution.

So, either $N\left(a\right)$or $N\left(b\right)=1$.

Thus 7 is irreducible.

Therefore, factorization of $7=7$

(c)Step 3: Factor  4+3i  as product of irreducibles in ℤ[i]

Let $4+3i=ab$

$N(4+3i)=25$

Now $N\left(a\right)|25$so $N\left(a\right)=5$.

Let $a=x+iy$then $N(x+iy)=x^2+y^2=5$.

This has solutions $x=\pm 1,y=\pm 2$ or $x=\pm 2,y=\pm 1$.

On testing these values, it shows

$4+3i=(1+2i)(2-i)$.

These factors are further irreducible because their norm is 5 which is prime.

(d)Step 4: Factor 11+7i  as product of irreducibles in ℤ[i]

Let $11+7i=ab$

$\begin{array}{c}N(11+7i)=121+49\\ =170\end{array}$

Now $N\left(a\right)|170$, so $N\left(a\right)=2,5,17$

Let $a=x+iy$ so $N\left(a\right)=x^2+y^2=2,5,17$

Solving separately, we get

$11+7i=(1+i)(1+2i)(1-4i)$

The factors are further irreducible because they have prime norms.