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Let $D$ be a principal ideal domain and $a$ and $b$ be nonzero elements of $D$. Prove that there exist elements $s$ and $t$ in $D$ such that $\gcd(a, b) = as + bt$.

I would like to use some properties of $\text{PID}$s to prove this but I am only thinking of well-ordering principle that is used to prove for integers, which I don't think I can use since $D$ is not necessarily the set of integers, right? Any ideas?

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    $\begingroup$ Consider the ideal generated by $a$ and $b$. $\endgroup$ – Lord Shark the Unknown Oct 30 '18 at 3:21
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Consider the ideal

$\langle a, b \rangle \subset D; \tag 1$

since $D$ is a principal ideal domain, we have $d \in D$ such that

$\langle a, b \rangle = \langle d \rangle; \tag 2$

this in itself is sufficient for

$\exists s, t \in D, \; d = as + bt; \tag 3$

now (2) implies

$d \mid a, \; d \mid b, \tag 4$

and if

$c \mid a, \; c \mid b, \tag 5$

then

$\exists x, y \in D \mid a = cx, \; b = cy; \tag 6$

inserting these equations into (3) yields

$d = as + bt = cxs + cyt = c(xs + yt), \tag 7$

whence

$c \mid d; \tag 8$

$d$ is thus a divisor of $a$ and $b$ which is itself divided by any $c$ such that (5) binds; but this is the definition of a greatest common divisor; therefore,

$d = \gcd(a, b). \tag 9$

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    $\begingroup$ What a beautiful reply! Truly easy to understand, thank you! $\endgroup$ – numericalorange Oct 30 '18 at 3:49
  • $\begingroup$ @numericalorange: thanks. It is really the "standard proof". And thanks for the "acceptance"! Cheers! $\endgroup$ – Robert Lewis Oct 30 '18 at 3:50
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Hint $\,\ c\mid\gcd(a,b)\!\iff\! c\mid a,b\!\iff\! (c)\supseteq (a),(b)\!\iff\! (c)\supseteq \overbrace{(a,b)=(d)}^{\Large as+bt\ =\ d\ }\!\iff\! c\mid d$

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  • $\begingroup$ Damn you are so concise, !!! $\endgroup$ – Robert Lewis Oct 30 '18 at 3:53

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