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I am trying to figure out whether $\mathbb{Z}[\sqrt{-11}]$ is a unique factorization domain. A hint provided is to "Consider for example the irreducibility of 2 and the primality of the ideal (2)". I assume that we will have to use the fact that in UFD's, irreducible elements generate (nonzero) prime ideals, and that the converse holds too.

So I first want to check whether 2 is irreducible and I set : $2=(a+b\sqrt{-11})(c+d\sqrt{-11})=(ac-11bd)+(ad+bc)\sqrt{-11}$ which implies that: $2=ac-11bd$ and $0=ad+bc$.

From there, I don't exactly see how to proceed in a smarter way than just trying out numbers for a,b,c and d.

Then I want to check whether the ideal generated by 2 is prime: An element in $(2)$ is of the form $2(x+y\sqrt{-11})$. Let two elements in our ring be such that their product is in $(2)$. We want to show that at least one of these elements must be in $(2)$ for the ideal to be prime.

$(p+q\sqrt{-11})(r+s\sqrt{-11})=2(x+y\sqrt{-11})$ implies that : $pr-11sq=2x \:\text{and}\: ps+qr=2y$.

Again, I don't see how to proceed from there. I'd be grateful for some insight on this!

Ava Rose
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  • Duplicate. See https://math.stackexchange.com/q/70976. –  Jun 20 '18 at 17:40
  • Alternatively, we have two irreducible factorizations for $12$: $$ \left(1+\sqrt{-11}\right)\left(1-\sqrt{-11}\right) = 12 = 3 \cdot 4 $$

    $1\pm\sqrt{-11}$ is irreducible because there is no element of norm $2,3,4,6$, the only possible nontrivial factors of $N(1\pm\sqrt{-11})=12$.

    $3$ is irreducible because there is no element of norm $3$, the only possible nontrivial factor of $N(3)=9$.

    $4$ is irreducible because there is no element of norm $2,4,8$, the only nontrivial factors of $N(4)=16$.

    They are not associates because they have different norms.

     – lhf Jun 20 '18 at 18:23

2 Answers2

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To see that $2$ is irreducible you can use the norm inherited from $\mathbb{Q}\left(\sqrt{-11}\right)$. We have that $N(2) = 4$ and if $2$ isn't irreducible we must have that $\mathbb{Z}\left[\sqrt{-11}\right]$ has an element of norm $2$. This means that $a^2+11b^2 = 2$ has a solution in integers, which is not true. Hence $2$ is irreducible.

Now if the ring is UFD we must have that $2$ is a prime too and so we have that $(2)$ is a prime ideal. However you have that $\left(1+\sqrt{-11}\right)\left(1-\sqrt{-11}\right) \in (2)$, but neither of them is inside the ideal, which contradicts the primality. Hence the ring isn't UFD.

Stefan4024
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This ring is not a U.F.D. because U.F.D.s are integrally closed, and as $-11\equiv 1\mod 4$, the integral closure of $\mathbf Q(\sqrt{-11})$ is $$\mathbf Z\biggl[\dfrac{1+\sqrt{-11}}2\biggr].$$

Bernard
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  • Thank you for your answer. However I haven't encountered integrally closed domains in class yet, so I'm looking for another way of proving it – Ava Rose Jun 20 '18 at 17:54