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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.

Alphonse
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  • I believe it is flat, judging from the last criterion on the wikipedia page, but unfortunately I cannot write up an answer at the moment. Use that the image of any map $\Bbb{Z}[2i]^n\ \longrightarrow\ \Bbb{Z}[i]$ is always $\Bbb{Z}[2i]$ or $0$. – Servaes Nov 01 '18 at 20:33
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    One thing is that $\mathbb{Z}[2i]$ is Noetherian and thus "flat $\iff$ projective" for finitely generated case. – freakish Nov 01 '18 at 22:07
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    Note that $\Bbb{Z}[i]$ is flat over $\Bbb{Z}[2i]$ if and only if $\Bbb{Z}[i]{\mathfrak{p}}$ is flat over $\Bbb{Z}[2i]{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}\subset\Bbb{Z}[2i]$. This is obviously true for every prime $\mathfrak{p}$ not containing $2$ as then the two localizations are the same, hence it remains to check this for the unique prime $\mathfrak{p}=(2,2i)\subset\Bbb{Z}[2i]$ lying over $2$. – Servaes Nov 01 '18 at 22:23
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    Didn't think too much about it but is $2i \otimes 1 + 2 \otimes -i$ nonzero in $(2) \otimes \mathbb{Z}[i]$? If so flatness fails. – Badam Baplan Nov 02 '18 at 04:37
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    Yes, it is not flat since it is integral, and integral + flat implies faithfully flat (which you know that doesn't hold). – user26857 Nov 02 '18 at 06:05

1 Answers1

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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 $\mathbb{Z}[pi]$

Badam Baplan
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