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Let $I＝(a＋bx)$ be an nonzero ideal of quotient ring $R＝\mathbb Z[x]/(x^2＋7)$, here $(a＋bx)$'s $x$ means image of $x$ in quotient ring $\mathbb Z[x]/(x^2＋7)$.

Then, why $R/I$'s order is $a^2＋7b^2$?

　

I checked some example when $a,b$ is small, but I couldn't find logic applicable to general case.

I guess this may something to do with the concept, norm.

Pont
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  • Amusing detail: "$x$ means image of $x$ in quotient ring". – user26857 Aug 19 '21 at 09:32
  • Have you tried enumerating (listing) the elements of R/I for small a and b? So the number of distinct elements if we take bx = -a and x^2 = -7. I think that should point you in the right direction. – Tempestas Ludi Aug 19 '21 at 10:33
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    Here is plenty of material about the quotient rings of a related ring (modulo $x^2+1$ instead of $x^2+7$). That should get you started. – Jyrki Lahtonen Aug 19 '21 at 13:23

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First, we have $R\cong\Bbb Z[\sqrt{-7}]$. If we can find an $n\in\Bbb Z$ such that $$\Bbb Z/(n)=\Bbb Z[\sqrt{-7}]/(a+\sqrt{-7}b)$$ then we are done. This would show that the order of the quotient ring is $|n|$.

Define the projection map $$\pi:\Bbb Z\rightarrow\Bbb Z[\sqrt{-7}]/(a+\sqrt{-7}b)$$

  • The map is surjective. We can see this as follows: Since $\gcd(a,b)=1$ by Bezout's lemma there exists $c,d\in\Bbb Z$ such that $$ac+bd=1$$ Suppose that $x+y\sqrt{-7}\in\Bbb Z[\sqrt{-7}]/(a+b\sqrt{-7})$, then we can write $$x+y\sqrt{-7}=x+y\sqrt{-7}-y(c+d\sqrt{-7})(a+b\sqrt{-7})\\=x-yac+7ybd$$ That is, $$\pi(x-yac+7ybd)=x+y\sqrt{-7}$$

  • $\ker(\pi)=(a^2+7b^2)$. Notice that $\ker(\pi)=\Bbb Z\cap (a+b\sqrt{-7})=(a^2+7b^2)$

It follows that $|n|=a^2+7b^2$.

cansomeonehelpmeout
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  • Could you tell me why Z∩(a＋b√7)＝(a^2＋7b^2)? And why we can assume a and b are coprime without loss of generality? – Pont Aug 19 '21 at 23:08