Non-maximal orders contains non-invertible ideals Let $K$ be an algebraic number field and $L$ the algebraic integers in $K$. Assuming an order $O$ is defined as a subring of $L$ that is of finite index as a subgroup of the additive group $L$, I will like to prove that if $O \neq L$ then $O$ has a non-zero ideal which is not invertible.
I already proved that $O$ is never integrally closed, which implies is not a Dedekind domain. Consequently, it contains a non-invertible non-zero ideal. However, I would like to see a constructive proof. Is there a way to actually compute this non-invertible ideal?
 A: Your idea of writing $L$ for the algebraic integers of a number field $K$ is terrible! I find such notation so bizarre that I refuse to follow it: $\mathcal O_K$ will be the ring of integers of $K$ in this answer.
You ask for a constructive proof that $O$ contains a nonzero noninvertible ideal, but you never told us how $O$ is created. What if $O$ is not constructive? You surely can't expect a constructive proof about ideals in $O$ if $O$ is not constructive.
A formula for a nonzero noninvertible ideal is easy. Since the order $O$ is non-maximal, the index $m = [\mathcal O_K:O]$ is finite and greater than $1$. The set $m\mathcal O_K$ is an example of a nonzero ideal in $O$ that is not invertible in $O$.  
To explain that, first note that the additive quotient group $\mathcal O_K/O$ has size $m$, so every element of this quotient group added to itself $m$ times vanishes in the quotient group, which means $m\mathcal O_K \subset O$. Obviously $m\mathcal O_K$ is an additive subgroup of $O$. Since $O \subset \mathcal O_K$, $m\mathcal O_K$ is carried back to itself under multiplication by each element of $O$, so $m\mathcal O_K$ is an ideal in $O$ and it's not $(0)$ since it contain $m$ (or since $m$ times a nonzero element of $\mathcal O_K$ is not $0$). Thus $m\mathcal O_K$ is an ideal of $\mathcal O_K$ and an ideal of $O$.
The following theorem implies $m\mathcal O_K$ is not invertible as a fractional $O$-ideal. 
Theorem: A nonzero ideal $\mathfrak b$ of $O$ that is also an ideal in $\mathcal O_K$ is not invertible as a fractional $O$-ideal.
Proof: If $\mathfrak b$ is invertible as a fractional $O$-ideal, then $$\{x \in K : x\mathfrak b \subset \mathfrak b\} = O.$$ 
However, since $\mathfrak b$ is a nonzero ideal in $\mathcal O_K$ and all fractional $\mathcal O_K$-ideals are invertible as fractional $\mathcal O_K$-ideals, $\{x \in K : x\mathfrak b \subset \mathfrak b\} = \mathcal O_K$.  Therefore $O = \mathcal O_K$, which is a contradiction.
This argument basically comes from Remark 3.3 of the file here.
