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Let $K = \mathbb{Q}[i]$ and $\mathcal{O}_K = \mathbb{Z}[i]$. Suppose $a = 1 + 2i$ and $q = 7$. Since $a$ and $q$ are coprime algebraic integer, then there must exists $b \in \mathcal{O}_K $ s.t. $a * b = 1 \pmod{q} $.

Is there an algorithm for finding that for a general number field $K$?

4 Answers 4

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Algorithm would be a bit exaggerated for a simple observation. In general, $z=u+iv$ in a unit mod $7$ in $\mathbf Z[i]$ if and only if its norm $N(z)=u^2+v^2$ is a unit in $\mathbf Z/7\mathbf Z$, and its inverse is $N(z)^{-1}\bar z$.

Hence, all you have to do is proceed with the extended Euclidean algorithm in $\mathbf Z$ for $N(z)$.

In the present case, $N(1+2i)=5$, its inverse mod. $7$ is $3$, hence $$(1+2i)^{-1} \bmod 7=3(1-2i)=3+i.$$

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    Nice idea (+1), but this seems to work only for $q\in\Bbb Z$. – 2017-01-02
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    For an irreducible of $\mathbf Z[i]\smallsetminus\mathbf Z$, you can use the extended Euclidean algorithm in this Euclidean domain. – 2017-01-02
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    yes, because the ring of Gaussian integers ia an euclidean domain. But these are very exceptional, the OP is asking for a "general" algorithm. – 2017-01-02
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    It is not clear to me the O.P. specifies a modulus in the number field. It could be meaningful only in the case of a ring of algebraic integers equipped with some Euclidean stathm. – 2017-01-02
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If I understood correctly, you need

$$(1+2i)(a+bi)=1+7m\;,\;\;a,b,m\in\Bbb Z\implies a-2b+(2a+b)i=1+7m\iff$$

$$\begin{cases}a-2b=1+7m\\{}\\2a+b=0\end{cases}\;\;\;\implies5a=1+7m\implies a=3\pmod 7\;,\;\;b=-2a=1\pmod7$$

Thus, for example, with $\;3+i\;$:

$$(3+i)(1+2i)=1+7i=1\pmod7$$

But you could pick as well $\;10+8i\;$ say , since

$$(10+8i)(1+2i)=-6+28i=1\pmod7$$

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    No, $q$ is assumed to be any element in $\mathbb{O}_K$ – 2017-01-02
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    @KunXianXia Yes, of course: **you wrote** $\;q=7\;$ , an algebraic integer...What's the problem? – 2017-01-02
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    I'm seeking an algorithm which takes two coprime element $a, q \in \mathbb{O}_K$ as input and outputs the inverse of $a$ mod $q$. – 2017-01-02
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    In other words, this algorithm can be seen as generalization of extended Euclidean algorithm for the $\mathbb{Z}$ case. – 2017-01-02
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    In *the case* you gave there is, and above is the proof. In general is a way more complex question as the $\;\mathcal O_K\;$ may not be a UFD – 2017-01-02
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    @KunXianXia Would it not stand to reason, then, that you'd want to ensure your ring to be a Euclidean domain? – 2017-01-02
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No. Consider $K=\Bbb Q[\sqrt{-5}]$. Let $a=1+\sqrt{-5}$, $q=2$.

Suppose that there exists $b,c\in \Bbb Z[\sqrt{-5}]$ such that $ab+cq=1$. But now note that $$3q=(1-\sqrt{-5})a$$ that is, $$3bq=(1-\sqrt{-5})ab=(1-\sqrt{-5})(1-cq)$$ therefore, $$q(3b+(1-\sqrt{-5})c)=1-\sqrt{-5}$$ which is clearly a contradiction.

In general, we can say that if the equation $ax\equiv 1\pmod m$ is solvable on some integral domain $R$ when $a$ and $m$ are coprime, then $R$ is an Unique Factorization Domain. But rings like $\Bbb Z[\sqrt d]$ where $d$ is a non square integer are not always UFD.

For those such rings that are Euclidean domains, the Euclidean algorithm works and you can use it to find inverses, just like in $\Bbb Z$.

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    Actually the very notion of $\rm gcd$ makes sense only in a PID – 2017-01-02
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    @AdLibitum Or, since we are working with fields of numbers, we can give the word 'greatest' (from greatest common divisor) the sense "greatest absolute value". But of course then the property of distinguished common divisor ($d\mid a\wedge d\mid b\implies d\mid\gcd(a,b)$) does not hold. Nonetheless, when I wrote $\gcd(a,m)=1$ I meant "$a$ and $m$ are coprime", and the notion of coprimality does not need $\gcd$. I have fixed it to avoid confusion. – 2017-01-02
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    @ajotatxe, the example you gave is not appropriate since the ideal $\langle a, q \rangle$ is a prime ideal not $\mathbb{O}_K$ itself – 2017-01-02
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In the specific case I would proceed as follows.

Since $7\equiv 3\bmod 4$ one knows that $7{\cal O}_K$ is a prime ideal and $$ {\cal O}_K/7{\cal O}_K\simeq{\Bbb F}_7[i] $$ is the field with $7^2$ elements.

Then one can solve $(1+2i)(x+2y)=1$ in ${\Bbb F}_7[i]$ which gives rise to a linear system in $\Bbb F_7^2$ pretty much as in DonAntonio's answer (which came out as I'm writing this). It gives the solution $$ x+2y\equiv 3-6i\equiv 3+i\bmod 7{\cal O}_K. $$

I guess that one can tackle the general problem along the same lines as long as the ideal $q{\cal O}_K$ is maximal.

If not, one has to preliminary find the decomposition $q{\cal O}_K=\prod_i\wp_i^{e_i}$ in ${\cal O}_K$ where the $\wp_i$ are maximal. Then your starting element $a$ will be invertible modulo $q$ if and only if it is invertible modulo each and every $\wp_i^{e_i}$ by the Chinese Remainder Theorem.

Of course in the rare event where ${\cal O}_K$ is euclidean, one can use the division algorithm to explicit the Bezout identity.