Let $\mathcal{O}(\mathbb{C}^n)_0$ denote the local ring of germs at the origin of holomorphic functions on $\mathbb{C}^n$. Consider the obvious map $$ \mathcal{O}(\mathbb{C}^n)_0 \otimes_{\mathbb{C}} \mathcal{O}(\mathbb{C}^m)_0 \rightarrow \mathcal{O}(\mathbb{C}^{n+m})_0. $$ Is it known whether or not this is a flat map of commutive $\mathbb{C}$-algebras?
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1$\begingroup$ By Tag 00HJ in Stacks Project, and the fact that for any Noetherian local ring the map $R\to\hat{R}$ is faithully flat, it suffices to check your statement on completions. The map induced on completions is an iso, hence flat. $\endgroup$– ChrisCommented Jan 1, 2023 at 12:00
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2$\begingroup$ I'm aware that flatness of maps of Noetherian local rings is detected by passing to completions; however I'm quite sure the algebra $\mathcal{O}(\mathbb{C}^n)\otimes_{\mathbb{C}} \mathcal{O}(\mathbb{C}^m)$ is only local for $m$ or $n$ equal 0. Of course, since localization is flat and preserves Noetherianness, this wouldn't be an issue if we knew the tensor product was Noetherian, but I don't see this. Is it just Weierstrass preparation? $\endgroup$– Pelle SteffensCommented Jan 1, 2023 at 12:43
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$\begingroup$ You're right, I took for granted that LHS was again a Noetherian local ring. I can't see at the moment why it's Noetherian $\endgroup$– ChrisCommented Jan 1, 2023 at 15:32
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