I am studying the preliminary part of a course in Complex Geometry, and it deals with analytic sets and holomorphic functions, at the germ level. At some point, it says that the ideals $I_k:=(z_1,\dotsc,z_k)\leq\mathcal{O}_{\mathbb{C}^n,0}$ (which is the ring of germs of holomorphic functions at 0 in $\mathbb{C}^n$) are all prime in $\mathcal{O}_{\mathbb{C}^n,0}$. I googled, maybe not too smartly, but I found nothing. I tried to prove this myself, but I got stuck at some point. How would I go about proving it?


MickG the self-answerer is back :). This time, I was asking a genuine question, I was showing my work… and I finished the proof! So here is what I typed in the question box before moving it here.

Here is what I did. Suppose $fg\in I$. We must show $f\in I\vee g\in I$. $fg$ is a germ of a holomorphic function, hence we can write it as a "multivariate power series", and being in $I$ it has the form:

$$fg=\sum a_{i_1,\dotsc,i_m}z_{i_1}^{m_1}z_{i_m}^{m_m},$$

where $i_i\leq k$ for all $i$ and the $a_{i_1,\dotsc,i_m}$ are holomorphic functions. I guess this leads me pretty much nowhere. I mean, I somehow thought I could reduce the $a_{i_1,\dotsc,i_k}$ to being constant, but I guess I cannot. The only thing I seem[ed] to be able to tell is that $fg$ cannot have a known term, that is, all the terms in its power series must contain at least a coordinate function with a strictly positive exponent. Anyways, in my mistake, I expanded $f,g$ as power serieses and split them into $f_1+f_2+f_3$ where $f_1$ only had terms of the form $cz_{i_1}^{m_1}\dots z_{i_m}^{m_m}$ with $i_i\leq k$, $f_2$ only had terms of the form $cz_{i_1}^{m_1}\dots z_{i_m}^{m_m}$ with $i_i>k$ plus maybe a constant term, and $f_3$ had mixed terms. Now then I had:


since $f_1,g_1\in I$ for sure. Then again, looking at it, $f_3,g_3$ can also be written in the form given for $fg$, since they only contain mixed terms: for each term, one separates the terms with the coordinate functions with index less than $k$ and puts the rest as a holomorphic function, so that all those terms are in $I$. So in the end:


$fg\in I$, hence $f_2g_2\in I$, but that doesn't contain any $z_i$ with $i\leq k$ with a strictly positive exponent, so none of its terms can be in $I$, hence to be i $I$ it must be zero. But then $f_2=0\vee g_2=0$, and this means, respectively, $f\in I\vee g\in I$. Whohoo, I thought I hadn't done it, and I just finished it! :)

So this is the last item of my "Share your knowledge, Q&A style!" tetralogy (or maybe pentalogy), the others being about the Spectral Theorem, Weak Compactness of Closed Unit Balls in Hilbert Spaces, and the Fundamental Theorem of Integral Calculus (and perhaps the Kuratowski-Ryll Nardzewski theorem, though I'm not sure I used the

Answer your own question - share your knowledge, Q&A style"

checkbox for this last cited post).

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