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?


3 Answers 3


Notice that: $(-1+2\sqrt{3})(1+2\sqrt{3})=11\in I$ Also $(1+2\sqrt{3})\sqrt{3}=6+\sqrt{3}\in I$. Thus $a+b\sqrt{3}+I=a+b(-6)+I=a-6b+I=[(a-6b)\mod\,11]+I$. Hence: $$\mathbb{Z}[\sqrt{3}]/I=\{a+I:0\leq a\leq 10\}$$

Now Consider $\phi:\mathbb{Z}_{11}\rightarrow \mathbb{Z}[\sqrt{3}]/I$ that sends $x$ to $x+I$. Verify that $\phi$ is a surjective group homomorphism. Thus, $|\mathbb{Z}[\sqrt{3}]/I|$ divides $11$ ......

Finally, show that $|\mathbb{Z}[\sqrt{3}]/I|\not=1$ by showing that $1\not\in I$ using a norm argument.

  • $\begingroup$ Would you mind giving some details as to why $a-6b+I=(a-6b)$ mod $(11+I)$ ? $\endgroup$
    – Phil-ZXX
    May 8, 2013 at 0:14
  • $\begingroup$ @Thomas The notation looks weird I will edit it $\endgroup$
    – Amr
    May 8, 2013 at 0:14
  • $\begingroup$ @Tthomas Is this better now ? $\endgroup$
    – Amr
    May 8, 2013 at 0:15
  • $\begingroup$ @Tthomas $a-6b=11k+(a-6b)\mod\,11$. Since $11\in I$, therefore $11k\in I$. Tthus, $(a-6b)\mod\,11+I=a-6b+I$ $\endgroup$
    – Amr
    May 8, 2013 at 0:16
  • $\begingroup$ Is this because 11 is the smallest positive integer in $I$? $\endgroup$
    – Phil-ZXX
    May 8, 2013 at 0:22

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

$Q \cong \mathbb{F}_{11}[x]/(x^2-3,2 \cdot 6 + 2 \cdot x) = \mathbb{F}_{11}[x]/(x^2-3,6+x)=\mathbb{F}_{11}/(5^2-3)=\mathbb{F}_{11}.$


In questions like this it can help to know the order of the quotient ring a priori. For this, one can use the following argument:

  • One has the chain of ideals $(11) = (1+2\sqrt{3})(1-2\sqrt{3}) \subset (1+2\sqrt{3}) \subset \mathbb Z[\sqrt{3}].$

  • Multiplication by $(1+2\sqrt{3})$ induces an isomorphism between $\mathbb Z[\sqrt{3}]/(1-2\sqrt{3})$ and $(1+ 2\sqrt{3})/(11)$.

  • Galois conjugation (swapping $\sqrt{3}$ and $-\sqrt{3}$) gives an isomorphism betweem $\mathbb Z[\sqrt{3}]/(1-2\sqrt{3})$ and $\mathbb Z[\sqrt{3}]/(1+2\sqrt{3}).$

Putting all this together, we find that the order of $\mathbb Z[\sqrt{3}]/(11)$ is equal to the square of the order of $\mathbb Z[\sqrt{3}]/(1+2\sqrt{3}).$ Since the former is isomorphic (as an abelian group) to $\mathbb Z/(11) \times \mathbb Z/(11)$, it has order $11^2$, and so $\mathbb Z[\sqrt{3}]/(1+2\sqrt{3})$ has order $11$.

In this particular case, we are done, since the quotient ring has order $11$ which is prime, and so is necessarily the prime field of order $11$.

In general, one can show that if $a$ and $b$ are coprime, then $\mathbb Z[\sqrt{3}]/(a + b \sqrt{3})$ is isomorphic to $\mathbb Z/(a^2 - 3b^2)$, and it is a good exercise to write down the details, using the same argument as in Amr's answer (combined with the obvious generalization of the above argument, to see that order of the quotient ring is equal to $|a^2 - 3b^2|$).

If you want to see another phrasing of essentially the same argument (in the context of the ring $\mathbb Z[i]$, but it goes exactly the same way) see this answer.


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