Show that the ideal $I = (2,1+\sqrt{-13})$ is a maximal ideal in $\mathbb{Z}[\sqrt{-13}]$? Is it principal?

To show that the ideal is maximal, I know that $I \text{ maximal} \iff \mathbb{Z}[\sqrt{-13}]/I \text{ is a field}\iff \text{only ideals of } \mathbb{Z}[\sqrt{-13}]/I \text{ are } 0,\mathbb{Z}[\sqrt{-13}]/I$.

I have determined that $I = \{(2a+c-13d) + (2b+c+d)\sqrt{-13} : a,b,c,d \in\mathbb{Z}\}$.

I now do not know how to continue showing that I is maximal.

To show $I$ is not principal, lets assume $I = (a)$ for some ring element $a$. Then since $2\in(a)$, we have $N(a) | 4$ so $N(a) = 1,2,4$. But $x^{2}+13y^{2} =1$ means $a=1\implies I = \mathbb{Z[\sqrt{-13}]}$ which is a contradiction since $1\not\in I$. $N(a) = 2$ has no solutions. So $N(a) = 4 \implies N(a)|13$ which is another contradiction (since $\sqrt{-13} \in I$)

  • $\begingroup$ It goes the same way as in this question. $\endgroup$ – Dietrich Burde Jan 4 '18 at 19:22
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    $\begingroup$ $\sqrt{-13}$ is not an element of the ideal. But $2$ and $1+\sqrt{-13}$ are. We have $N(2)=4$ and $N(1+\sqrt{-13})=14$. So if $a$ generates $I$, then $N(a)$ must be a factor of both $4$ and $14$.... $\endgroup$ – Jyrki Lahtonen Jan 4 '18 at 19:24
  • $\begingroup$ @JyrkiLahtonen oops, you're right, and the only factor of $4$ and $14$, for which $N(a)$ has a solution is $1$ and $N(a)=1$ leads to a contradiction $\endgroup$ – pureundergrad Jan 4 '18 at 19:28
  • $\begingroup$ $\gcd(4,14)=2$ but that doesn't really help the search for $a$ :-) $\endgroup$ – Jyrki Lahtonen Jan 4 '18 at 19:29

Use your first equivalence: $I$ is maximal if and only if ${\mathbb Z}[\sqrt{-13}]/I$ is a field. The trick now is to rewrite the quotient ring until you see that it's a field: \begin{equation*} {\mathbb Z}[\sqrt{-13}]/(2, 1 + \sqrt{-13}) \cong {\mathbb Z}[x]/(2, 1 + x, x^2 + 13) \cong {\mathbb F}_2[x]/(1 + x, x^2 + 1) \cong {\mathbb F}_2. \end{equation*} Each of the isomorphisms requires justification, but every time it's the 'obvious' map that gives the isomorphism and it's the same type of reasoning for each of this type of problems.

  • $\begingroup$ I cannot understand why any of those equivalences are true $\endgroup$ – pureundergrad Jan 4 '18 at 19:19
  • $\begingroup$ first quotient by $2$ then map $x=1$ using $x+1=0$, then set $1^2=-13\implies 14=0$ which imposes no further conditions $\endgroup$ – operatorerror Jan 4 '18 at 19:41

Let $I\subset J$ be a larger ideal and $a+b\sqrt{-13}\in J-I$ since $2\in I$ we may assume that $a,b =\pm 1$. Now $\pm(1\pm\sqrt{-13})\in I$, so this leaves the possibilities $\pm 1$ and $\pm \sqrt{-13}$ both of which entail that $1\in J$ and so $J$ is not proper.

For non principal, if $2=xy$ then as you observe $4=Nx Ny$ and so $Nx=1$, or $Ny=1$ or $Nx=Ny=2$. The latter is impossible and the former implies that one of $x$ or $y$ is in $\mathbb{Z}$, also impossible.


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