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I have an integral domain $(R,+,\cdot)$ and I need to show that $x \in R$ is a unit iff it is a divisor of every $a \in R$.

As I just began to study integral domains I do not really know where to start. Do you have any tips?

jjbinks
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  • If $x\in R^$ and $a\in R$, then $a=x(x^{-1}a)$. If $x\in R$ is a divisor of every element, then in particular is a divisor of $1$. Then there is $y$ such that $1=xy$. So, $x\in R^$ – plop Jan 07 '23 at 17:22
  • Check what you have written. It does not match the title. You are wishing to prove that $x$ is an element of the integral domain iff it is a divisor of every $a$? Or you are wishing to prove that $x$ is a unit iff it is a divisor of every $a$? – JMoravitz Jan 07 '23 at 17:22
  • Sorry you are right, I edited the question. – jjbinks Jan 07 '23 at 17:32
  • Units $u$ are the divisors of $1$. So $u\mid 1\mid r\Rightarrow u\mid r$ for all $r\in R$ by transitivity of divides (cf. linked dupe). Conversely $u$ divides all $r \Rightarrow u\mid 1\Rightarrow u$ a unit. – Bill Dubuque Jan 07 '23 at 18:48

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The term “divisor” leads to thinking in terms of multiplication inverse. $$d\ne 0:\frac{a}{d}=a \Leftrightarrow a\cdot \frac{1}{d}=a\Leftrightarrow a\cdot (d)^{-1}=a\Leftrightarrow (d)^{-1}=1\Leftrightarrow d=1^{-1}=1(\therefore)$$

WindSoul
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