# Less discriminating discriminants

Suppose that I want to know whether a polynomial $P(z)$ has a root with multiplicity at least three. This is obviously an algebraic condition, but is there some reasonably concise set of conditions defining the variety (in the space of coefficients)? This must have been studied by the ancients. It is clearly necessary that the discriminant vanish, and also that the resultant of the polynomial $P$ and the second derivative $P^{\prime\prime}$ vanish, but, just as obviously, not sufficient...

EDIT Abhinav certainly gives a nice answer to the question, but the question I would REALLY like to know the answer to is: what is the degree of the variety as a function of the partition (as in @Gjergji's answer). Maybe I should read the reference...

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Er. Why is the condition you describe not sufficient? – Qiaochu Yuan Aug 7 '11 at 1:50
@Qiaochu: just because $P$ and $P^\prime$ have a common root, and $P$ and $P^{\prime\prime}$ have a common root, does not mean that all three do, unless I am VERY confused. – Igor Rivin Aug 7 '11 at 1:54
@Igor, see the end of page 5 of the paper I linked to. The formula for the degree goes back to Hilbert. – Gjergji Zaimi Aug 7 '11 at 2:53
@Gjergji: Ah, thanks! – Igor Rivin Aug 7 '11 at 14:08

Here's one way to think of this. Let $\lambda$ be a new variable and let $Q(z) = \lambda P'(z) + P''(z)$. Let $R(\lambda)$ be the resultant of $P$ and $Q$. Then we want $R$ to identically vanish (i.e. want to see if there's value of $z$ for which $P$ and $Q$ have a common zero, for every value of $\lambda$). So the coefficients of $R$ will describe an ideal cutting out the variety you want. This argument should work at least in characteristic $0$.

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That's a nice argument, though the system you get is a little scary... – Igor Rivin Aug 7 '11 at 2:24

The situation becomes considerable harder beyond testing for a root of multiplicity $2$. The problem itself is very old. A recent article about this is "On equations defining coincident root loci" by J. Chipalkatti. The result is that the algebraic conditions can be expressed through a cohomology group of a certain complex of $SL_2$ representations

Note: This is addressing the general question of when a polynomial has roots of multiplicities indexed by a partition $\lambda$. The question itself is about $\lambda=(3,1^{n-3})$.

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For partitions of the special form $\lambda=(k,1,1,\dots,1)$ (i.e., hook shapes) there is a very explicit description of the ideal of definition of the corresponding set of polynomials (i.e., those polynomials with a root of order $\geq k$). Namely, it is the span of the set of (symmetric) Jack polynomials $f_\mu$, with parameter specialized to $-1/k$, and where $\mu$ ranges over all partitions satisfying

$$\mu_i-\mu_{i+k-1} \geq 2$$ for all $i$. This is the main result of the paper http://arxiv.org/pdf/math/0112127 by Feigin, Jimbo, Miwa, and Mukhin.

For the application to degree, one might note that this computes the Hilbert function of this ideal quite explicitly in combinatorial terms, though for your problem you are probably interested in the grading where each elementary symmetric function has degree one, and I don't know, off the top of my head, how to extract that information. Maybe I'll post a separate question so we can see if anywhere else here knows.

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Remark: For the application to degree, it's probably useful that this is really the whole ideal of definition, and not just an ideal cutting the variety out set theoretically. – Stephen Aug 8 '11 at 18:36
By the way, the paper arxiv.org/pdf/math/0404079 considers a somewhat more general version of the problem and gives some partial results for some other $\lambda$'s. – Stephen Aug 8 '11 at 18:48