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Well I want to prove that $\overline{11}$ is an unit in $\Bbb Z[i]/<5+6i>$ and that it is also the inverse of itself.

Please check if it is mathematically correct:

An element is an unit, if it is coprime to $5+6i$, so Calculate $ N(5 + 6i) $:

$N(5 + 6i) = 5^2 + 6^2 = 25 + 36 = 61$

Calculate $( N(11) ):$ Since $( 11 )$ is a real integer, we treat it as $( 11 + 0i )$, so

$[N(11) = 11^2 = 121.]$

Check if $( N(11) )$ and $( N(5 + 6i) )$ are coprime: We now check if $( N(11) = 121 )$ and $( N(5 + 6i) = 61 )$ are coprime. Since $( 61 )$ is a prime number and does not divide $( 121 = 11^2 )$, we conclude that $( 61 )$ and $( 121 )$ are coprime.

Therefore, $ 11 $ is coprime to $ 5 + 6i $, which implies that $( \overline{11} )$ is a unit in $( \mathbb{Z}[i]/\langle 5 + 6i \rangle )$.

Now to prove that it is the inverse of itself: Let $a \in \overline{11}$ need to find $x$ such that $$ax \equiv1 (mod 11)$$ which gives as $0,1,2,3,4,5,6,7,8,9,10$

gagamaga
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  • Note that $11\times11=2\times61-1$, so $11\times11\equiv-1\bmod61$, so $11\times(61-11)\equiv1\bmod61$, so $11\times50\equiv1\bmod61$ – J. W. Tanner Nov 03 '24 at 20:44
  • $\overline{11}$ is the inverse of itself because $\overline{11}\times\overline{11}=121\times121\equiv-1\times-1=1\pmod{61}$ and therefore $\pmod{5+6i}$ – J. W. Tanner Nov 03 '24 at 21:11
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    Is $\overline x$ the same as $N(x)$ in your notation? That's confusing, especially since $\overline x$ often denotes the residue class – J. W. Tanner Nov 03 '24 at 21:19
  • Use \langle and \rangle for delimiters, not < and >. – Arturo Magidin Nov 03 '24 at 21:27
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    @J.W.Tanner I meant the residue class ! :) – gagamaga Nov 03 '24 at 21:42
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque Nov 03 '24 at 21:43
  • If $\overline{11}$ denotes residue class, then $\overline{11}$ is not its own inverse, because $\overline{11}\times\overline{11}=121\equiv-1\not\equiv1\pmod{61}$ – J. W. Tanner Nov 03 '24 at 21:46
  • By the dupe: $11$ and $\alpha = 5!+!6i$ have coprime norms $\Rightarrow (11)+(\alpha)=(1),$ in $,\Bbb Z[i],$ so $,(11)=(1)$ in $\Bbb ,\Bbb Z[i]/(\alpha).\ \ $ – Bill Dubuque Nov 03 '24 at 21:52
  • @BillDubuque: OP also wanted proof that $\overline{11}$ is the inverse of itself – J. W. Tanner Nov 03 '24 at 22:14
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    @J.W.T Dupe targets titled question (posts should have only one question). But - of course - inverting $11$ mod $61$ is also a big dupe $|\ \ $ – Bill Dubuque Nov 03 '24 at 22:25
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    @J.W.T More conceptually $,11^2\equiv -1,$ by $,11\equiv -i\pmod{!5!+!6i}$. Indeed scaling $,-6i\equiv 5,$ by $(-6)^{-1}\equiv 10,$ yields $,i\equiv 50\equiv -11\pmod{!61}.,$ Said in equivalent ideal language $,(5+6i) = (5+6i,61) = (11+i,61),,$ a special case of Hermite normal form reduction (as here). $\ \ $ – Bill Dubuque Nov 03 '24 at 22:59

1 Answers1

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Your proof is correct, but the last paragraph is unclear and wrongly phrased. To prove that $\overline{11}$ is the inverse of itself, it is enough to solve for $11^2 = 121 \equiv u \pmod{5+6i}$ (where $u = \pm 1 \ {\rm or} \ \pm i$). One can verify that for $u=1$, no solution exists, but for $u= -1$, a solution exists; hence $11$ is its own inverse $\pmod{5+6i}$.

Edit: $(a,b) = (10, -12)$ is a solution for $121 - (-1) = (5+6i)(a+ib)$.

Anuradha N.
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  • Isn't $121\equiv\color{red}-1\pmod{5+6i}$? (Cf. my comment to OP) – J. W. Tanner Nov 03 '24 at 20:56
  • I think OP wants $\overline{11}\times\overline{11}=121\times121\equiv-1\times-1=1\pmod{5+6i}$ – J. W. Tanner Nov 03 '24 at 21:08
  • Your solution is simpler and more direct. My answer involves solving for a pair of linear equations in two integer variables. – Anuradha N. Nov 03 '24 at 21:18
  • I still think what you say is incorrect. $11$ is not its own inverse, though $\overline{11}=121$ is its own inverse $\pmod{5+6i}$ – J. W. Tanner Nov 03 '24 at 21:22
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    I have added an edit to clarify. I assume he means residue class of $11$ (and not its norm). – Anuradha N. Nov 03 '24 at 21:29
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    See OP's definition of $\overline{x}$ in the title (and my comment to OP that it's confusing) – J. W. Tanner Nov 03 '24 at 21:31
  • Please strive not to post more (dupe) answers to PSQs, off-topic SVs, dupes of FAQs This is enforced site policy, see here. – Bill Dubuque Nov 03 '24 at 21:45
  • @J.W.T $\ \bar r := r+I,$ is common notation for an equivalence (congruence) class in a quotient ring $,R/I,,$ i.e. $,R\bmod I,,$ so it is not clear why you found it confusing. The bar notation is never used for norms (but it is also used for conjugates). $\ \ $ – Bill Dubuque Nov 04 '24 at 02:26
  • @BillDubuque: the initial version of OP had "Let ... $\overline{x}$ be the modulus of $x=a+bi∈\mathbb Z[i]$" in the title; that is why I found it confusing (and asked for clarification in a comment to OP, q.v. above) – J. W. Tanner Nov 04 '24 at 03:51