MHB 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 $$2$$ is irreducible in $$\mathbb{Z} + \mathbb{Z} \sqrt{ -5 }$$, it must be the case that $$\alpha \sim 1$$ or $$\alpha \sim 2$$. ... ...
My question is as follows ... how does $$2$$ being irreducible imply that $$\alpha \sim 1$$ or $$\alpha \sim 2$$. ... ...?
Hope someone can help ...

Peter============================================================================NOTEThe notation $$\alpha \sim 1$$ is Alaca and Williams notation for $$\alpha$$ and $$1$$ 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