I would like to know how to think about primary ideal geometrically. Vaguely speaking, I think it's an irreducible closed subscheme with some "infinitesimal" data - however, I am not sure how to make this precise.

In particular, when I try to play with this example $\mathbb{C}[x,y,z]/(xy-z^2)$ (from Atiyah-Macdonald), it is easy to check that $(x,z)^2$ is not primary even though it is really the ruling $(x,z)$ together with some "infinitesimal" data. I am guessing the geometric obstruction lies in the non-smoothness at the origin, which makes $(x,z)^2$ not a "real" infinitesimal neighborhood around the origin, but again I don't know how to make this precise.

Another example is $(x^2,xy) \subset \mathbb{C}[x,y]$. Is there a conceptual explanation that this is not a primary ideal by drawing pictures?

Any insight is welcome. Thanks!

(p.s. I already read Professor Emerton's answer here Geometric meaning of primary decomposition, but I don't think it addresses my question above.)


For a primary ideal $I$, if we take the radical $\sqrt I$, we get a prime ideal $\mathfrak p$ which corresponds to the underlying irreducible component $Z=V(I)$. Moreover we have some "infinitesimal"/non-reduced data along $Z$. Being primary says that this non-reduced data must be "uniformly" supported over all of $V(I)$ and not on some proper subvarieties.

Precisely this means if we have a non-reduced element $a$ (or any element for that matter), then for any subvariety $W\subsetneq Z$ with corresponding prime ideal $\mathfrak q\supsetneq\mathfrak p$, $a$ cannot be killed by $\mathfrak q$ (since $\mathfrak q$ are the functions vanishing on $W$, $\mathfrak q$ killing $a$ would mean that $a$ is supported on $W$). This gives the equivalent definition for $I$ being primary (when $A$ is noetherian): $A/I$ having only one associated prime.

In the example of $I=(x^2,xy)\subset k[x,y]$, geometrically we are looking at the intersection of the "double $y$-axis" $(x^2)$ and a "cross" $(xy)$: this gives the $y$-axis with some non-reduced structure in the $x$-direction at the origin. As $\sqrt I=(x)$, the underlying variety $Z$ is the $y$-axis. The function $x$ is zero along the $y$-axis, except at the origin $W$, where it need to vanish with order 2. But then it can be killed by some function defining $W$ in $Z$, say $y$. So we have $x\notin I$, $y\notin \sqrt{I}$, but $xy\in I$.

Similarly, in the example $I=(x,z)^2\subset k[x,y,z]/(xy-z^2)$, we intersect the cone $xy=z^2$ with the "double $y$-axis": we get "uniform" reduced structure in the $z$-direction along the whole $y$-axis, but with extra reduced structure in the $x$-direction at the origin. So again we can take the function $x$, which is supported at the origin and killed by $y$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy