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Let $A$ be a principal ideal domain and $S = \{a_1,...,a_n \}$ a subset of $A$ and let $(S)$ be the ideal generated by the elements of $S$. Define $T_S = \{t \in A \ \vert \ \forall i =1,...,n, t \mid a_i \}$.

Show that $(S) = A \Leftrightarrow T_S \subset A^\times$, where $A^\times$ denotes the group of units.

$\Leftarrow:$ If $T_S \subset A^\times$ then for $t \in T_S$, we have that $t \in A^\times$. Thus $t$ is invertible and from the definition of $T_S$ we know that $t \vert a_i, \forall i$. From here I don't know how to proceed.

user26857
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    Please do not delete your question immediately after receiving an answer. This is rude to the person who took the time to answer your question, and rude to potential future readers who might benefit from the question and answer pair. – Xander Henderson Apr 30 '21 at 15:57

2 Answers2

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Let the ideal generated by $S$ be given by $(f)$ for some $f \in A$.

Note that $f \in T_S$ so if $T_S \subseteq A^\times$, then $f$ is a unit so $(f) = (1) = A$. Conversely, if $\langle S\rangle = (1)$, then we can write $1 = \sum_{i = 1}^n c_i a_i$ by the definition of the ideal generated by $S$. Hence, if some $t$ divides all $a_i$, it divides every $c_i a_i$ so that $t$ divides the sum, which is 1. Hence, $t$ is a unit so $T_S \subseteq A^\times$.

user26857
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Daniel
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  • Note that the proof practically writes itself if we use the innate universal proiperties - as I explain in my answer. – Bill Dubuque May 06 '21 at 08:35
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Conceptually it can be viewed as the special coprime case of the general fact that PIDs are gcd domains, whose ideals are generated by any gcd of their generators, since

$$\begin{align} \overbrace{(a_1,\,\ldots,\,a_n)}^{\large (a_1)\,+\cdots +\, (a_n)} &\,=\, D,\ \ \ \text{wich means, by def'n of ideal sum, that}\\[.2em] \iff\ \ \ C\supseteq (a_1),..,(a_n) &\iff \, C\,\supseteq\, D,\ \ \ \text{for all ideals $\,C\subseteq A$}\\[.2em] \iff\ \ \ (c) \!\supseteq \!(a_1),..,(a_n) &\iff (c)\supseteq (d)\ \ \ \text{by $A$ is a PID} \\[.2em] \iff\ \ \ \ \ c\ \ \mid\ \ a_1,\ldots,a_n\,\ &\iff\ \ c\ \ \mid\ \ d,\,\ \ \ \text{by contains = divides in a PID}\\[.2em] \iff\ \ \ \ \ \ \ \gcd(a_1,\cdots, a_n)\! &\ \ \sim\ d,\ \ \ \text{where $\,\sim\,$ means associate} \end{align}\qquad$$

where the final $\!\iff\!$ uses the gcd universal property, and the first is the universal property of the ideal sum $\,C\supseteq A_1+A_2\!\!\iff\! C\supseteq A_1,A_2,\, $ where $(a_1,\ldots,a_n) = (a_1)+\cdots (a_n)\,$ by definition.

As is often the case, the proof is straightforward - practically writing itself - when we employ the basic universal properties that define the objects.

For motivation see here for the case when the PID $=\Bbb Z$.

Bill Dubuque
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