Is $\Bbb Z[i]$ flat as $\Bbb Z[2i]$-module? Is $\Bbb Z[i]$ flat as $\Bbb Z[2i]$-module ? (Here $i^2 = -1$).
I know it's flat as a $\Bbb Z$-module, since it is torsion-free and $\Bbb Z$ is Dedekind. Actually it is even free as $\Bbb Z$-module. But I don't think that $\Bbb Z[i]$ is free over $\Bbb Z[2i]$.
I think that it is not faithfully flat : here, we have an ideal $I \subset \Bbb Z[2i]$ such that $I \neq I^{ec}$. But I don't know if this is because it is not flat or not faithful.
 A: Just to close this up and generalize a little bit... using user26857's approach in the comments, we can show that $\mathbb{Z}[i]$ is not flat over $\mathbb{Z}[pi]$ for any prime $p \in \mathbb{Z}$.  

Lemma $\mathbb{Z}[i]$ is integral over $\mathbb{Z}[pi]$.  

Proof: Let $z = a + ib \in \mathbb{Z}[i]$ with $a,b \in \mathbb{Z}$. Consider the imaginary component of $(a + ib)^p$. Using the binomial formula, we get an expression for it as $\sum\limits_{k=1 \\ \text{k odd}}^p (-1)^{\lfloor k/2\rfloor}a^{p-k}b^k{p \choose k}$.  If $p$ is odd, this reduces to $(-1)^{\lfloor p/2\rfloor} b^p \text{ mod } p$ (since ${p \choose k}$ is divisible by $p$ for $k \not= 0, p$).  From Fermat's little theorem, we see that $z^p - (-1)^{\lfloor p/2 \rfloor}z \in \mathbb{Z}[pi]$, so that $z$ is integral over $\mathbb{Z}[pi]$.  Otherwise $p = 2$ and we have $z^2 \in \mathbb{Z}[2i]$.  

Lemma: A flat integral extension is faithfully flat.

Proof: Integral extensions satisfy lying over for primes, so the induced map on spectra will be surjective.  In the presence of flatness, this is equivalent to faithful flatness.  

Lemma: $\mathbb{Z}[i]$ is not faithfully flat over $\mathbb{Z}[pi]$

Proof: As in your linked question, we note that faithfully flat extension are such that ideals are preserved by extension-contraction. We establish that $(p) \subset \mathbb{Z}[pi]$ is such that $(p)^{ec} \not= (p)$ by noting that $pi \in (p)^{ec}$ but $pi \notin (p)$

Conclusion.  For any prime $p \in \mathbb{Z}$, $\mathbb{Z}[i]$ is not flat over $\mathbb{Z}[pi]$

