How can I prove an ideal is a product of two irreducible ones I'm trying to solve this question:

I have a guess that $(6+\sqrt{11})=(2,4+\sqrt{11})(2,-3\sqrt{11})$ using some formulas in this book page 48. However I couldn't verify if the multiplication of these ideals are indeed equal to $(6+\sqrt{11})$ and prove they are primes.
If I'm wrong, what are the standard techniques I can use to solve this question?
I really need help.
Thanks in advance. 
 A: Hints: 


*

*$(10+3\sqrt{11})(10-3\sqrt{11})=1$.

*$(4-\sqrt{11})^2(10+3\sqrt{11})=?$


Do interpret this in term of prime ideals and units.
A: How you might approach it in a systematic way:
If $K = \mathbb{Q}(\sqrt{11})$, the norm of $(6+\sqrt{11})$ is $36-11 = 25$ so we know that $(6+\sqrt{11})$ is a product of prime ideals with norm dividing $5$. In general we need to look at how each rational prime dividing the norm of the element we're interested in ramifies in $K$, so here we need only examine $5$. 
$11 = 3 \pmod{4}$ so we know that $\mathcal{O}_K = \mathbb{Z}[\sqrt{11}]$. Then we can apply Dedekind's criterion to the polynomial $f(X)=X^2-11$ to factorise the ideal of any rational prime $p$, in particular, $p = 5$. (Why is this? Under what conditions can we apply Dedekind's criterion?)
Thus $\bar{f}(X)=X^2-1=(X-1)(X+1)\pmod{5}$ so $(5)=(5,1+\sqrt{11})(5,1-\sqrt{11})=P_1P_2$, say. For ideals in $\mathcal{O}_K$ for any number field $K$, $I|J$ is the same as $J \subset I$, so $P_2$ cannot divide $(6+\sqrt{11})$ else it would contain $2$, which is impossible since each prime ideal contains a unique rational prime. 
This is a very general method, and I would say that understanding how it works gives a very useful tool in factorising ideals in many number fields.
