Fractional Ideal Inverse

See the proof that this is an inverse below for some motivation on why we use this definition.

Here, $a\mathbb{Z}$ is the set of $\mathbb{Z}$ multiples of $a$. The values in $F$ that we can multiply this by in order to stay within the ring are fractions of the form $\frac{b}{a}$ for some $b\in \mathbb{Z}$. That way, the product $a\times \frac{b}{a}=b\in \mathbb{Z}$.

From this we establish some important basic properties of the inverse which are essential for it being used as a multiplicative inverse in the fractional ideal group.

Proof

Our definition for the fractional ideal inverse is very carefully constructed such that we have

by definition (stare at the above and think carefully to see how deliberately constructed this is).

Let $r\in R$, and $i\in I$ with $i\ne 0$. Then, we have that $\frac{r}{i}\in I^{-1}$ since $\frac{r}{i}\times i=r\in R$. As such, we have that $r\in I^{-1}I$.

The fact that $I{I}^{-1}=I^{-1}I$ follows simply from the fact that $R$ is an integral domain, and hence commutative, a property that the ideal multiplication inherits.

Proof

First note that $R^{-1}=\{x\in F:xR\subseteq R\}\subseteq R$, and if $r\in R$ then $r\in \{x\in F:xR\subseteq R\}$ since $1\in R$, and therefore clearly $R^{-1}=R$ (this is just a consequence of the group structure we can construct, but we prove it in a more elementary way here so that this makes sense before that result).

Assume now that $I\subseteq R$. If $r\in R$, then clearly $r\in F$, and $rI\subseteq R$ since $I\subset R$ and $r\in R$, so $rI$ is a set of products of elements in $R$. As such $r\in I^{-1}$ and so more generally $R\subseteq I^{-1}$.

We get the strict inequality by applying $R^{-1}=R$ above.