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Background:

Theorem 6.6 Let $I$ be an ideal in a ring $R$ and let $a,c\in R,$ Then $a\equiv c \pmod I$ if and only if $a+I=c+I.$

Theorem 10.8 Every Euclidean domain is a principal ideal domain.

Exercise 15: Let $I$ be a nonzero ideal ini $\mathbb{Z}[i]$, show that the quotient ring $\mathbb{Z}[i]/I$ is finite.

Proof 1: By Theorem 10.8 $I=(b)$ for some nonzero $b.$ If $a\in \mathbb{Z}[i],$ then $a=bq+r$ with $r=0$ or $\delta(r)<\delta(b),$ and, hence, $a\equiv r \pmod I.$ By Theorem 6.6, the number of distincct cosets of $I$ (congruence classes mod $I$) is at most the number of possible $r$'s under division by $b.$ Show that there are only finitely many possible $r$'s.

Proof 2: Say $I=\langle a+bi \rangle.$ Then $a^2+b^2+I=(a+bi)(a-bi)+I=I$ and therefore $a^2 +b^2\in I.$ For any $c,d\in \mathbb{Z},$ let $c=q_1(a^2+b^2)+r_1$ and $d=q_2(a^2+b^2)+r_2,$ where $0\leq r_1,r_2<a^2+b^2.$ Then $c+di+I=r_1+r_2i+I.$

Questions:

I have a few related questions for the two proofs above to the quoted exercise. In proof 1, I don't understand when it says "the number of distinct cosets of $I$ (congruence classes mod $I$) is at most the number of possible $r$'s under division by $b.$" Is the author saying that I am suppose to count the number of equivalence classes? Because it asks the reader to show that there can only be finitely many $r$s. But in proof 2, I am not sure why it need to use two separate integers $c,d$ with respect to $(a^2+b^2)$ in order to use the division algorith twice to show that there are two remainders $r_1, r_2$ and that they are used to satisfy Theorem 6.6. How is that showing there are only two $r'$s? I mean why only two integers, or does it mean that any more would imply that the extra ones would go into one of the two equivalence classes $[c],[d]$? (here i am using the notation $[c]$ to denote equivalence classes). Also when this question came up in the text, the notion of norm for gaussian integers has not been introduced yet.

Thank you in advance

Arturo Magidin
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Seth
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1 Answers1

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If $a + I$ and $c + I$ are distinct cosets, then $a \not\equiv c \pmod{I}$ by the first theorem. Since an element is equivalent to its remainder modulo $I$, we have $a \equiv r_a \pmod{I}$ and $c \equiv r_c \pmod{I}$, implying that $r_a \not\equiv r_c \pmod{I}$.

This means the map taking an element $x$ to its remainder $r_x$ is injective. The number of possible remainders is finite, since we must have $\delta(r) < \delta(b)$, and the number of elements of $\mathbb{Z}[i]$ with norm less than $\delta(b)$ is finite.

The second proof is completely different. It says that the norm of $a+bi$ is an element of $I$, so one may perform Euclidean division with $a^2+b^2$ to write any element as $r_1 + r_2i$ where $0 \leq r_1, r_2 < a^2+b^2$.

For example, if $I = \langle 3+ 4i \rangle$, then $25 \in I$. Then if you gave me an arbitrary coset, like $(102 + 264i) + I$, I would simply tell you that $102 + 264i \equiv 2 + 14i \pmod{I}$, and hence the coset you gave me is actually equivalent to $(2 + 14i) + I$.

There would then be at most (this is overestimating) $25 \times 25$ possible cosets.

Orange Mushroom
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  • @OrangeMusroom for the first proof, when you said "the number of elements of $\mathbb{Z}[i]$ with norm less than $\delta(b)$ is finite." basically we don't care what the $\delta$ is suppose to be, only that $\delta(b)$ is finite. Is that correct? – Seth Jan 18 '24 at 09:20
  • for the second proof,, $102 + 264i \equiv 2 + 14i \pmod{I}$ means that $102 + 264i - (2 + 14i) + I=100+250i+I\in I$ but since the norm of $\langle 3+ 4i \rangle$ is $25$, then $25\mid 100, 25\mid 250$, but how do you get $25\times 25$ cosets? – Seth Jan 18 '24 at 09:24