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I am now taking undergraduate abstract algebra and the professor's proof for Bezout's Identity at class goes in this way:

Proposition: Let a ring $R$ be a PID and $\gcd(a,b)=1$ in $R$, then $\exists \alpha, \beta \in R, \alpha a + \beta b = 1$.

Proof:

Let $I = \langle a,b\rangle$ be an ideal generated by $a$ and $b$, i.e. $I = \{au + bv \mid u,v\in R \}$. Since $R$ is a PID, then $\exists x \in R$, such that $I = \langle a, b\rangle = \langle x\rangle = \{xy \mid y \in R \}$.

Claim: $x$ is a unit, because: $(a \in I \land b \in I) \Rightarrow (x \mid a \land x \mid b)$. Since $\gcd(a,b) =1 $, we have $x$ is a unit.

Then, since $x \in \langle a,b\rangle$, we have $\exists \alpha', \beta' \in R$, $x = \alpha' a + \beta' b$, and since $x$ is unit, we have $xx' = 1$ for some $x' \in R$. So $1 = xx' = (x' \alpha') a + (x' \beta') b = \alpha a + \beta b$, where $\alpha = x' \alpha'$, $\beta = x' \beta'$.

Question: My question is about Claim part. I know why we have $a \in I$ and $b \in I$ and why we have $x \mid a$ and $x \mid b$, but I don't understand why we have $x$ is a unit here.

Hope someone can help me with this. Thanks!

ZYX
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  • If there is any mistake in my classnote, please point it out and I will edit it! Thanks! – ZYX Nov 04 '22 at 01:36

1 Answers1

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You've shown that $x$ must satisfy both $x\mid a$ and $x\mid b$. In other words, $x$ is a common divisor of $a$ and $b$. By definition of gcd (the greatest common divisor), it must then hold that $x\mid \gcd(a,b)$, in which case $x\mid 1$. This automatically implies that $x$ is a unit (i.e., $\exists y$ such that $xy = 1$).

Rob
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