Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

up vote 8 down vote accepted

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.

share|improve this answer
    
Would you mind giving some details as to why $a-6b+I=(a-6b)$ mod $(11+I)$ ? –  Tom May 8 '13 at 0:14
    
@Thomas The notation looks weird I will edit it –  Amr May 8 '13 at 0:14
    
@Tthomas Is this better now ? –  Amr May 8 '13 at 0:15
    
@Tthomas $a-6b=11k+(a-6b)\mod\,11$. Since $11\in I$, therefore $11k\in I$. Tthus, $(a-6b)\mod\,11+I=a-6b+I$ –  Amr May 8 '13 at 0:16
    
Is this because 11 is the smallest positive integer in $I$? –  Tom May 8 '13 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

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

share|improve this answer

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 $Z[\sqrt{3}]/(1-2\sqrt{3})$ 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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.