Argument and motivation behind maps used in finding inrreducibles in an integral domain

Question

I present the solution to a past question that I didn't had time to understand even on a superficial level.

I am unable to understand the classical argument employed in the methods for finding irreducibles. I have stared at this for 5 whole hours.

Here's what I know: 21=7.3=$(1+2\sqrt{-5})(1-2\sqrt{-5})$

The factors 7,3, $(1+2\sqrt{-5}) and (1-2\sqrt{-5})$ are all non-unit and clearly are non-zero. They satisfy ONE condition for the definition for an element/ factor to be an irreducible.

To ensure the factors are irreducibles, we need to ensure that there exists two elements in the Integral domain $\mathbb{Z}\left [ \sqrt{-5} \right ]$ such that a factor, say 3 = xy where x,y are elements in $\mathbb{Z}\left [ \sqrt{-5} \right ]$. It suffice for us to ensure that either x or y are units.

But in the solution, the argument assumes that x and y are non-units. Is this a proof by contradiction? Also, why it it necessary for a map N commonly used in determining primes, irreducibles and associates to exists? I'm extremely confused when authors toggle between the image of the elements under the map N and the elements itself in the integral domain.

Answer 1

Yes, it is a proof by contradiction. An element $z ∈ ℤ[\sqrt{-5}]$ is reducible if there are nonunits $x, y ∈ ℤ[\sqrt{-5}]$ such that $z = xy$.

If, for any pair of non-units $x, y ∈ ℤ[\sqrt{-5}]$, it is impossible that $z = xy$, then $z$ must be irreducible.

This is what is done in the quoted solution for $z = 3$, $z = 7$ and $z = 1 ± 2\sqrt{-5}$: They assume that there are nonuints $x, y ∈ ℤ[\sqrt{-5}]$ with $z = xy$ and find a contradiction.

To reach the contradiction, they throw $N$ at the assumption $z = xy$ to get $N(z) = N(x)N(y)$. This is an equation in $ℤ$ where we know unique prime factorisation is possible, so it’s easier to reason here. This why this is done.

Such a neat map $N \colon A → ℤ$ doesn’t exist for general integral domains $A$. Here, in this case, you can think of it as coming from the square of the absolute value $ℂ → ℤ, z ↦ |z|^2$. So we use that the domain in question $A$ is actually embedded in a ring we already know a lot about: $ℂ$.

(More generally, the $N$ stands for norm. Whenever $L / K$ is a field extension, any element $a ∈ L$ acts as a $K$-linear map on $L$ by right-multplication $(~·a) \colon L → L,~x ↦ xa$, so it makes sense to look at its determinant – which is called its norm: $N(a) = \det (~·a)$.)

Answer 2

Suppose that the map $N$ is $\rm\color{#c00}{multiplicative}$ $\,N(xy) = (Nx)(Ny),\,$ and, furthermore, that $\,\color{#0a0}{x\mid Nx}.\,$ Then $\,Na$ irreducible $\,\Rightarrow\, a$ irreducible, since

$\ \ a = bc\,\color{#c00}\Rightarrow\, Na = Nb\, Nc,\ $ so $\,Nb\,$ or $\,Nc\,$ is a unit, say $Nc.\ $ Then $\,\color{#0a0}{c\mid Nc}\mid 1\,$ so $\,c\,$ is unit.

Yours is special case $\,Na = a\bar a = {\rm Norm}(a)$

Remark $ $ The key idea is that multiplicative maps preserve multiplicative properties (which is brought to the fore in divisor theory). Above we pullback the "atom" (irreducible) property along the multiplicative norm map $\,N$.

In fact much of the multiplicative structure of a number ring is reflected in its monoid of norms. For example, in many favorable contexts (e.g. Galois) a number ring enjoys unique factorization iff its monoid of norms does. See this answer for further discussion, including literature references (Bumby and Dade, Lettl, Coykendall).