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I am reading the concept of complex spaces from "Several Complex Variables VII: Sheaf-Theoretical Methods in Complex Analysis" By Grauert, Peternell and Remmert. They defined a point $x$ in a complex space $(X,\mathcal{O}_{X})$ to be irreducible if $\mathcal{O}_{X,x}$ is an integral domain.

Now, $x$ has a neighbourhood $U$ in $X$ s.t. $(U,\mathcal{O}_{U})$ is isomorphic (as ringed spaces) to the complex model space(P-26 of the same book mentioned above for precise definition) defined by holomorphic functions $f_1,\ldots,f_{k}\in\mathcal{O}(D)$, where $D$ is an open subset in $\mathbb{C}^{n}$, for some $n$. We'll denote the model space as $V(f_{1},\ldots,f_{k})$, which equals $(Z(f_{1},\ldots,f_{k}),\mathcal{O}_{D}/\mathcal{I}_{D}|_{Z(f_{1},\ldots,f_{k})})$. Here $Z(f_{1},\ldots,f_{k})$ is the common zero set of $f_{1},\ldots,f_{k}$ and $\mathcal{I}_{D}$ is the ideal sheaf of $\mathcal{O}_{D}$ given by $\mathcal{I}_{D}:=\mathcal{O}_{D}f_{1}+\ldots+\mathcal{O}_{D}f_{k}$. Finally, assume that $(\phi,\widetilde{\phi})$ is the isomorphism between the ringed spaces mentioned above.

In the context of complex analytic variety there is another notion of irreducible points given as follows: A point $p$ in a complex analytic variety $X$ is irreducible if the analytic germ $X_{p}$ can't be written as the union of two analytic germs $X_{1p},X_{2p}$ with the property that $X_{ip}\neq X_{p}$, for $i=1,2$.(For the definition of analytic germ, see Definition 1.1.22, P-18, "Complex Geometry" by Daniel Huybrechts). From this, I think the following is true: A point $x$ in the complex space $(X,\mathcal{O}_{X})$ is irreducible iff $Z(f_{1},\ldots,f_{k})_{\phi(x)}$ is an irreducible analytic germ.

Forward Part: Suppose $x$ is irreducible. From the morphism $(\phi,\widetilde{\phi})$ one can get that $\mathcal{O}_{X,x}\simeq\mathcal{O}_{D,\phi(x)}/\mathcal{I}_{D,\phi(x)}$. Consequently, $\mathcal{O}_{D,\phi(x)}/\mathcal{I}_{D,\phi(x)}$ is an integral domain or equivalently, $\mathcal{I}_{D,\phi(x)}$ is a prime ideal of $\mathcal{O}_{D,\phi(x)}$. So $\sqrt{\mathcal{I}_{D,\phi(x)}}$ (radical of $\mathcal{I}_{D,\phi(x)}$) $=\mathcal{I}_{D,\phi(x)}$. Now, by Nullstellensatz $\sqrt{\mathcal{I}_{D,\phi(x)}}=I(Z(\mathcal{I}_{D,\phi(x)}))$(Proposition 1.1.29, P-19, "Complex Geometry" by Daniel Huybrechts), where $Z(\mathcal{I}_{D,\phi(x)}):=\bigcap_{\zeta\in\mathcal{I}_{D,\phi(x)}}Z(\zeta)=\bigcap_{i=1}^{k}Z(f_{i})_{\phi(x)}$ and for any analytic germ $X_{p}$ in $\mathbb{C}^{n}$, the ideal $I(X_{p}):=\{\eta\in\mathcal{O}_{\mathbb{C}^{n},p}| X_{p}\subseteq Z(\eta)\}$. Furthermore, Lemma 1.1.28, P-19 of the same book says that an analytic germ $X_{p}$ is irreducible iff $I(X_{p})$ is a prime ideal. Applying this to our case, we get that $Z(\mathcal{I}_{D,\phi(x)})=\bigcap_{i=1}^{k}Z(f_{i})_{\phi(x)}=Z(f_{1},\ldots,f_{k})_{\phi(x)}$ is an irreducible analytic germ.

Backward part(here I am stuck): Suppose $Z(f_{1},\ldots,f_{k})_{\phi(x)}$ is an irreducible analytic germ. This means $I(Z(\mathcal{I}_{D,\phi(x)}))=\sqrt{\mathcal{I}_{D,\phi(x)}}$ is a prime ideal of $\mathcal{O}_{D,\phi(x)}$. Now, if one can show that $\mathcal{I}_{D,\phi(x)}$ itself is a prime ideal of $\mathcal{O}_{D,\phi(x)}$, then $\mathcal{O}_{D,\phi(x)}/\mathcal{I}_{D,\phi(x)}$ (equivalently, $\mathcal{O}_{X,x}$) is an integral domain and we are done. But how does it follow from what we have that $\mathcal{I}_{D,\phi(x)}$ is a prime ideal$?$ Any help is appreciated.

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There is a difference between $\sqrt{\mathcal I_{D,\phi(x)}}$ and $\mathcal I_{D,\phi(x)}$ only if the germ is not reduced, and in this case, the two definitions are simply not equivalent. Otherwise, they are, and this is exactly what your proof shows.

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  • $\begingroup$ So, you are saying that the backward part doesn't hold necessarily because the hypothesis is not enough to ensure that $\phi(x)$ is a reduced point of $V(f_{1},\ldots,f_{k})$ (equivalently, $x$ is reduced point of $(X,\mathcal{O}_{X})$), let alone being an irreducible point which is far stronger and being a reduced point is same as getting the two above-mentioned ideals as equal, which we clearly don't have. $\endgroup$
    – Sam
    May 4, 2018 at 2:22
  • $\begingroup$ Indeed, classical examples of non-reduced points, for example, double points in $\mathbb{C}$ ($V(z^{2})$) will serve as a counter-example here. $\endgroup$
    – Sam
    May 4, 2018 at 2:29
  • $\begingroup$ Yes, @Sam, exactly. Note that in Huybrechts' definition, the germ is irreducible if and only if its reduction is irreducible. $\endgroup$
    – Ben
    May 4, 2018 at 8:18

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