Ideal of $\mathbb{Z}[\sqrt{10}]$: principal, prime or maximal?

Question

my task is following

Let $I=\{a+b\sqrt{10}:13\mid2a-b\}$ be a subset of the ring $\mathbb{Z}[\sqrt{10}]$. Decide if $I$ is an ideal of $\mathbb{Z}[\sqrt{10}]$ and if so, decide if it is principal, prime or maximal.

I've proved that $I$ is ideal indeed (addition is trivial and it is closed under multiplication because $13\mid2a-b\ \Leftrightarrow\ 13\mid a-20b$).

But I have problem with the properties of $I$. I know that $I$ is prime/maximal iff $\mathbb{Z}[\sqrt{10}]/I$ is an integral domain / a field. I don't know, how to continue -- $I$ is kernel of some ring homomorphism, but I have no idea how to use it. I suppose we would like to use something like $f:\mathbb{Z}[\sqrt{10}]\to\mathbb{Z}_{13},a+b\sqrt{10}\mapsto[2a-b]_{13}$ but is this even a map?... Edit: oh, of course $f$ is a map, since the element $a+b\sqrt{10}$ represents only itself...

@user26857 I have a hint, that I should use map $F:a+b\sqrt{10}\mapsto a^2-10b^2$. Suppose $I=(a+b\sqrt{10})$. If $c+d\sqrt{10}\in I$, then there is $x+y\sqrt{10}\in\mathbb{Z}[\sqrt{10}]$ such, that $c+d\sqrt{10}=(x+y\sqrt{10})(a+b\sqrt{10})$ and applying $F$ I should have $a^2-10b^2\mid c^2-10d^2$, right? — byk7, Nov 30 '16 at 14:31
Hint $\ 10\equiv 6^2\pmod{13}., $ See my answer if it is not clear where that leads. — Bill Dubuque, Nov 30 '16 at 15:05
@user26857 I don't know wheter $I$ is principle or not. So I suppose it is. (if it is principle indeed i want to find generator, if not I should obtain contradiction). Let be $I=(a+b\sqrt{10})$. If $c+d\sqrt{10}\in I$ then there are $x,y\in\mathbb{Z}$ such, that $c+d\sqrt{10}=(x+y\sqrt{10})(a+b\sqrt{10})$. Now I would like use $F:a+b\sqrt{10}\mapsto a^2-10b^2$ somehow. $F$ should be an analogy to similar situation, when we researched subsets of rings $\mathbb{Z}[i]$ and $\mathbb{Z}[i\sqrt{5}]$. We used there a map $g:\mathbb{C}\to[0,\infty),z\mapsto|z|^2$, so we obtained divisibility condition — byk7, Nov 30 '16 at 20:28
@user26857 Ok, so $I=\left(13,\sqrt{10}-6\right)=\left{(13x-6y)+y\sqrt{10}\ |\ x,y\in\mathbb{Z}\right}$. If $I=\left(a+b\sqrt{10}\right)$, then $a^2+10b^2\mid(13x-6y)^2+10y^2,\forall x,y\in\mathbb{Z}$? If so, could you explain me this a little bit more, please? It's not clear to me... — byk7, Nov 30 '16 at 22:05
@user26857 Yea, a typo. :) I came back to $x,y$ because then $\left|a^2-10b^2\right|\mid\left|(13x-6y)^2-10y^2\right|$, so I can generate candidates for the right side easily. E.g. for $x=0,y=1$ I have $\left|a^2-10b^2\right|\mid26$, for $x=1,y=0$ I have $\left|a^2-10b^2\right|\mid169$, hence $\left|a^2-10b^2\right|\mid13$. Or not? — byk7, Nov 30 '16 at 22:24

lhf · Accepted Answer · 2016-11-30T13:44:22.973

5

If you have proved that $I$ is an ideal, then you have proved that the map $$f:\mathbb{Z}[\sqrt{10}]\to\mathbb{Z}_{13} \mbox{ given by } a+b\sqrt{10}\mapsto[2a-b]_{13} $$ is a ring homomorphism with kernel $I$.

By considering the elements with $a=0$, it is clear that $f$ is surjective.

What does this tell you about $f$ ?

edited Nov 30 '16 at 13:44

answered Nov 30 '16 at 13:33

lhf

216,483

Something like there is a homomorphism g: Z[sqrt(10)] / I → Z/13Z, such that g∘p=f, where p is projection Z[sqrt(10)] → Z[sqrt(10)] / I, and g is surjection as well? – byk7 Nov 30 '16 at 13:40
1

@byk7: You're overthinking it a bit. What does the First Isomorphism Theorem let you conclude about $I,$ given that it is the kernel of the ring homomorphism $f$ mapping onto the field $\Bbb Z_{13}$? – Cameron Buie Nov 30 '16 at 13:48
1

@CameronBuie oh, you mean, that $\mathbb{Z}[\sqrt{10}] / I$ is isomorphic to $\mathbb{Z}_{13}$? Or that $I=\ker f$? – byk7 Nov 30 '16 at 13:52
2

@byk7: The isomorphism part. What does that tell you about $I$, since $\Bbb Z_{13}$ is a field? – Cameron Buie Nov 30 '16 at 13:58
2

@CameronBuie yea, just got it... since $\mathbb{Z}[\sqrt{10}]/I\cong\mathbb{Z}{13}$ and $\mathbb{Z}{13}$ is a field, then $I$ is prime as well as maximal... – byk7 Nov 30 '16 at 14:02

score 1 · Answer 2 · answered Nov 30 '16 at 15:03

1

Hint $\,\ 13\mid 2a\!-\!b\iff 13\mid 2(a\!+\!6b)\iff 13\mid a\!+\!6b\iff a+b\sqrt{10}\in (\sqrt{10}\!-\!6,\,13)$

answered Nov 30 '16 at 15:03

Bill Dubuque

272,048

ok, so there are at most two generators of $I$, but what to do now, please? – byk7 Dec 01 '16 at 23:52
@byk7 What part(s) are you stuck on? – Bill Dubuque Dec 02 '16 at 00:13
well, I have two generators of $I$, but this doesn't say, that $I$is not principal. I should use a norm $a+b\sqrt{10}\mapsto\left|a^2-10b^2\right|$, but dunno how. – byk7 Dec 02 '16 at 00:24
2

@byk7 Let $,w = 6!-!\sqrt{10}.,$ If $,(w,13)=(v),$ then $,v\mid w,$ so $,v'\mid w',$ so $,vv'\mid ww'=26.,$ Also $,v\mid 13,$ so $,v'\mid 13'=13,$ so $,vv'\mid 13^2,,$ so $,vv'\mid (26,13^2)=13(2,13)=13., $ Thus if $,v = a+b\sqrt{10},$ then $,vv'=a^2!-10b^2 = \pm 13.,$ But that has no integer solutions, being unsolvable mod $10.\ \ $ – Bill Dubuque Dec 02 '16 at 00:34
1

@byk7 Btw, proving the nontrivial direction $(\Leftarrow)$ of the final arrow in the answer is very simple via this criterion. – Bill Dubuque Dec 03 '16 at 22:40
Last thing -- why $v\mid w\Rightarrow v'\mid w'$ ? – byk7 Dec 07 '16 at 13:54
1

@byk7 $, w = vu,\Rightarrow,w' = (vu)' = v'u',$ since conjugation is a ring hom. – Bill Dubuque Dec 07 '16 at 14:32
@byk7 See also this answer where I show how to use the above mentioned criterion. – Bill Dubuque Dec 11 '16 at 18:12

Ideal of $\mathbb{Z}[\sqrt{10}]$: principal, prime or maximal?

2 Answers2

Linked