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If $\alpha$ is a quadratic integer in $\mathbb{Q}[\sqrt{d}]$Q, then define a notion of congruence $\pmod{\alpha}$. Furthermore, define $+$, $−$, and $\times$ for congruence classes, and show that this notion is well-defined.

I believe that the notion of congruence $\pmod{\alpha}$ means $x+y\sqrt{d}\equiv z+w\sqrt{d}$,
$x\equiv z$ and $y\equiv w$ modulo $\alpha$. Also, I was able to define $+$ for all congruence classes, which would be $[a,b]+[c,d]=[a+c,b+d]$.

What would the congruence classes be for $-$ and $\times$? Also, how would I be able to show that this notion is well-defined?

Arturo Magidin
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user687319
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  • Do you see what does the quotient ring $\Bbb{Z}[i]/(2+i) \Bbb{Z}[i]$ look like ? How many elements, characteristic ? – reuns Aug 19 '19 at 04:32
  • I'm sorry, I'm not sure I understand what you mean. – user687319 Aug 19 '19 at 05:26
  • @reuns Why is there a second $\Bbb Z[i]$? Is it part of the ideal? – gen-ℤ ready to perish Aug 19 '19 at 06:43
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    The way I would have stated @reuns’s question is: what does it mean for $z\equiv w\pmod{1+i}$ ? Here, $z,w$ are to be Gaussian integers, i.e. of form $m+ni$. Your proposed definition was fine in the case that $\alpha\in\Bbb Z$ . – Lubin Aug 19 '19 at 13:41
  • Congruence is defined for a commutative ring the same as in $,\Bbb Z,,$ i.e. $,a\equiv b\pmod {!c}\iff c\mid a-b,$ in $R,,$ i.e. $, a-b = cd,$ for some $,d\in R.,$ The proofs of the Congruence Sum & Product Rules are the same as in $,\Bbb Z,,$ which yields that addition & multiplication is well defined on the congruence classes. This will be algebraically reified when you study quotient rings, where $, R\bmod c,$ becomes $,R/cR\ \ $ – Bill Dubuque Aug 19 '19 at 13:42

1 Answers1

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Since subtraction is just addition in reverse, e.g., $x - y = x + (-y)$, you can focus your efforts on addition and multiplication.

The next step, I think, is to review what congruence means in $\mathbb Z$, since you're familiar with that domain of numbers, right?

Given plain integers $n$, $r$, $m$, what does it mean for $n \equiv r \pmod m$ to be true? It means that there is some integer $q$ such that $n = qm + r$. But a lot of times we don't really care what $q$ is.

Likewise in a quadratic integer ring. If $n$, $r$, $m$ are integers in a particular ring of $\mathbb Q(\sqrt d)$, $n \equiv r \pmod m$ means that $n = qm + r$, where $q$ is also an integer in $\mathcal O_{\mathbb Q(\sqrt d)}$, but we aren't terribly concerned about it for our purposes.

One of the commenters suggested looking at $\langle 2 + i \rangle$ in $\mathbb Z[i]$ and its cosets. I think $\langle 4 + \sqrt{14} \rangle$ in $\mathbb Z[\sqrt{14}]$ might make a better example for you.

If $a$ and $b$ are "plain" integers such that $(a - b \sqrt{14})(a + b \sqrt{14}) = N$ is an odd integer in $\mathbb Z$ (meaning that $N \equiv 1 \pmod 2$), I assert that $a \pm b \sqrt{14} \equiv 1 \pmod{4 + \sqrt{14}}$.

For example, $(7 - \sqrt{14})(7 + \sqrt{14}) = 35$. We see that $$7 - \sqrt{14} = (23 - 6 \sqrt{14})(4 + \sqrt{14}) + 1$$ and $$7 + \sqrt{14} = (5 - \sqrt{14})(4 + \sqrt{14}) + 1.$$ Looking at specific numbers should really help clarify these easy concepts that have been obscured by jargon.

Robert Soupe
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