Irreducible factors of number in $\Bbb Z[\sqrt{3}]$

Question

So I have to give a irreducible factor decomposition of $11$ in $\Bbb Z[\sqrt3]$. I know that I have to use factors of the form $(a+b\sqrt{3})$ but I don't know how to do it :(

As this seems like a homework assignment; are there any theorems or examples about such a ring in your literature so far? — Servaes, Jan 19 '17 at 09:10

InsideOut · Answer 1 · 2017-01-19T09:54:30.950

2

HINT: You are looking for two factors of the form $a+b\sqrt{3}$ and $c+d\sqrt{3}$ such that

$$(a+b\sqrt{3})(c+d\sqrt{3})=11.$$

Note that $11=11+0\sqrt{3}$, thus you have to solve a system like this:

\begin{cases} ac+3bd=11\\ ad+cb=0 \end{cases}

edited Jan 19 '17 at 09:54

answered Jan 19 '17 at 09:15

InsideOut

6,883

If the system haven't solution means that $11$ is irreducible. I am sure that it isn't a square. – InsideOut Jan 19 '17 at 09:18
This is what I needed! I didn't have any examples. Thanks. – Kiss-shot Jan 19 '17 at 09:20
Ok, so this system does not have a solution where a,b,c,d are integers. It means 11 is irreducible in $Z[\sqrt{3}]$, right? – Kiss-shot Jan 19 '17 at 09:27
1

In this case there is a solution of the system – InsideOut Jan 19 '17 at 09:30

Irreducible factors of number in $\Bbb Z[\sqrt{3}]$

1 Answers1