Units of $R[X]/(aX-1)$ 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?

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 $R[x]/I$.
