Irreducibles and Primes in Integral Domains ....

[Math Amateur]

I am reading "Introductory Algebraic Number Theory"by Saban Alaca and Kenneth S. Williams ... and am currently focused on Chapter 1: Integral Domains ...

I need some help with understanding Example 1.4.1 ...

Example 1.4.1 reads as follows:View attachment 6516
In the above text by Alaca and Williams we read the following:

"... ... From the first of these, as \(\displaystyle 2\) is irreducible in \(\displaystyle \mathbb{Z} + \mathbb{Z} \sqrt{ -5 }\), it must be the case that \(\displaystyle \alpha \sim 1\) or \(\displaystyle \alpha \sim 2\). ... ...
My question is as follows ... how does \(\displaystyle 2\) being irreducible imply that \(\displaystyle \alpha \sim 1\) or \(\displaystyle \alpha \sim 2\). ... ...?
Hope someone can help ...

Peter============================================================================NOTEThe notation \(\displaystyle \alpha \sim 1\) is Alaca and Williams notation for \(\displaystyle \alpha\) and \(\displaystyle 1\) being associates ...

Alaca's and Williams' definition of and properties of associates in an integral domain are as follows:https://www.physicsforums.com/attachments/6517

 

[vidyarth]

The meaning of an element of a ring being irreducible is, that it cannot be expressed as a product of two distinct elements in the ring upto units barring itself and unity, upto equivalence, i.e., its associates. Thus, since $2$ is irreducible in the ring $\mathbb{Z}+\mathbb{Z}\sqrt{5}$, therefore the only two elements which can divide $2$ in the ring are, $2$ and $1$, upto equivalence, which implies $\alpha\sim1$ or $\alpha\sim2$.

 

[Math Amateur]

The meaning of an element of a ring being irreducible is, that it cannot be expressed as a product of two distinct elements in the ring upto units barring itself and unity, upto equivalence, i.e., its associates. Thus, since $2$ is irreducible in the ring $\mathbb{Z}+\mathbb{Z}\sqrt{5}$, therefore the only two elements which can divide $2$ in the ring are, $2$ and $1$, upto equivalence, which implies $\alpha\sim1$ or $\alpha\sim2$.

Thanks vidyarth ... I appreciate your help ...

Peter