TITLE: What is the quotient $\mathbb Z[\sqrt{3}]/(1+2\sqrt{3})$?
QUESTION [13 upvotes]: I am currently doing a past paper and it asks the following:

Prove that for $I=(1+2\sqrt{3})$ we have $\mathbb Z[\sqrt{3}]/I$ a field with $11$ elements.

If I assume standard algebraic number theory then it would only be a few lines:
We know $N(I)=|\mathcal O/I|$ where $\mathcal O=\mathbb Z[\sqrt{3}]$. And the norm of a principal ideal is equal to the absolute value of the norm of its generator, i.e. $N(I)=|N(1+2\sqrt{3})|=|1-3\cdot 4|=11$. So $|\mathcal O/I|=11$ and hence $\mathcal O/I$ must be the finite field with $11$ elements.
But this paper is for a normal Commutative Algebra module and I cannot assume any of this. Is there any other way to approach this?

REPLY [5 votes]: There is a general method for computing these quotients, which is quite straight forward when you feel comfortable with polynomials.
Since $\mathbb{Z}[\sqrt{3}] \cong \mathbb{Z}[x]/(x^2-3)$, we have $Q:=\mathbb{Z}[\sqrt{3}]/(1+2 \sqrt{3}) \cong \mathbb{Z}[x]/(x^2-3,1+2x)$. In $Q$ we have $0=(2x+1)(2x-1)=4 x^2-1=11$. Therefore
$Q \cong \mathbb{F}_{11}[x]/(x^2-3,2 \cdot 6 + 2 \cdot x) = \mathbb{F}_{11}[x]/(x^2-3,6+x)=\mathbb{F}_{11}/(5^2-3)=\mathbb{F}_{11}.$