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I am reading Dummit & Foote's book and on page 282, I can't understand the part that I have offset and bolded in the following quote. How can we conclude that $ay-19bx=cq+r$? I know that there is different proof for the same question but I am curious to understand the proof of Dummit & Foote's book.

Example

Let $R = \mathbb{Z}\left [(1 + \sqrt{-19})/2 \right ]$ be the quadratic integer ring considered at the end of the previous section. we show that the positive field norm $N(a + b(1 + \sqrt{-19})/2) = a^2 + ab + 5b^2$ defined on $R$ is a Dedekind-Hasse norm, which by Proposition 9 and the results of the previous section will prove that $R$ is a P.I.D. but not a Euclidean Domain.

Suppose $\alpha, \beta$ are nonzero elements of $R$ and $\alpha/\beta \not \in R$. We must show that there are elements $s,t \in R$ with $0 < N(s \alpha - t \beta) < N(\beta)$, which by the multiplicativity of the field norm is equivalent to

$$ 0 < N \left ( \frac{\alpha}{\beta} s - t \right ) < 1. \quad \quad (*)$$

Write $\frac{\alpha}{\beta} = \frac{a + b \sqrt{-19}}{c} \in \mathbb{Q} \left [ \sqrt{-19} \right ]$ with integers $a,b,c$ having no common divisor and with $c > 1$ (since $\beta$ is assumed not to divide $\alpha$). Since $a,b,c$ have no common divisor there are integers $x,y,z$ with $ax + by + cz = 1$.

Write $ay - 19bx = cq + r$ for some quotient $q$ and remainder $r$ with $|r| \leq c/2$ and let $s = y + x \sqrt{-19}$ and $t = q - z \sqrt{-19}$.

Then a quick computation shows that

$$0 < N \left ( \frac{\alpha}{\beta} s - t \right ) = \frac{(ay - 19bx - cq)^2 + 19 (ax + by + cz)^2}{c^2} \leq \frac{1}{4} + \frac{19}{c^2}$$

and so $(*)$ is satisfied with this $s$ and $t$ provided $c \geq 5$.

Bill Dubuque
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Babak
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    I've edited your question to use mathjax (which is searcahble) rather than an image (which isn't) so that other users will have an easier time finding this question. In the future you should do the same ^_^ – Chris Grossack Mar 19 '23 at 03:28
  • All they are doing is applying the division algorithm. Divide $ay-19bx$ by $c$ to get a quotient $q$ and a remainder $r$. If $r$ is larger than $|c|/2$, then replace it with $r-c$ and replace $q$ with $q+1$. – Arturo Magidin Mar 19 '23 at 03:35
  • As in the dupe they choose the unique remainder $r$ in a balanced (signed) residue system $!\bmod c,,$ which is a complete system of residues $!\bmod c\ \ $ – Bill Dubuque Mar 19 '23 at 04:44
  • See also this answer. – Bill Dubuque Mar 19 '23 at 04:51

1 Answers1

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$ay - 19bx$ is some integer. Now we can use the division algorithm to write $ay - 19bx = qc + r$.

Typically we assume the remainder $r$ lies in $\{0, 1, 2, \ldots, c-1\}$, but we don't have to. We can take the remainder to land in any consecutive run of $c$ numbers. For instance, we can choose $\{-(c-1)/2, \ldots, -1, 0, 1, \ldots, (c-1)/2 \}$ (with an obvious modification if $c$ is even). As an example, instead of writing

$$17 = 5 \cdot 3 + 2$$

we could write

$$17 = 6 \cdot 3 - 1$$

If we do this, instead of knowing $0 \leq r < c$, we learn that $|r| \leq c/2$, as desired.

I hope this helps ^_^

Chris Grossack
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