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In the most upvoted answer to another question here, the author states that:

$R\to R[x]$ followed by the quotient map $R[x]\to R[x]/(ax-1)$. Call this $f$. Note that $f(a)$ is a unit in $R[x]/(ax-1)$

If I understand correctly, the result is $f(a)=ax^0\mod{ax-1}=ax^0$, and I fail to see why this is necessarily a unit, as $a\in R$ is just a non-zero element in an integral domain, so it doesn't necessarily have an inverse.

Why is $f(a)$ a unit? Am I calculating it wrong?

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Ruben Kruepper
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Let the ideal $(ax-1) = I$.

The map $f$ sends $a \mapsto a+I$ in the quotient ring and $(x+I)(a+I) = ax+I$ but $-ax+1 \in I$ so $(x+I)(a+I) = 1+I$.

Hence $f(a)$ is a unit in $R[x]/I$.

GoodThanks
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  • Could you please add some detail to the definition of the map $f$? I think I'm understanding it wrong. If $f$ sends $a$ to $a+I$, in the quotient Ring $a+I=a$ right? – Ruben Kruepper May 27 '20 at 16:49
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    Some people may write $a$ for $a+I$ in the quotient ring (I find this can be slightly misleading) but it is important to remember that in a quotient ring there may be many ways to write the same element. Really $a$ is an element of $R \leqslant R[x]$ but then the quotient map sends $a \mapsto a+I$ – GoodThanks May 27 '20 at 17:17