Coefficients of a polynomial also are the roots of the polynomial?

How many real solutions $(r_1, r_2, \cdots, r_n)$ are there such that $(r_1, r_2, \cdots, r_n)$ are the roots of the polynomials $x^{n} + r_1 x^{n-1} + r_2 x^{n-2} + \cdots + r_n$

For $n = 2, 3, 4$ I found $2, 4, 5$ real solutions, and $2, 6$, and according to WA and assuming a double root, $24$ complex solutions. Is it possible to generalize this for real/complex solutions? In particular, a proof (or dispute of the fact) that the number of complex solutions follows the factorials, and if possible also find a pattern for the number of real solutions. I would prefer not an answer that is just a computation to show a pattern up to some $n$.

Edited because of bounty.

• From Vieta's formulas, it follows that $\displaystyle\prod_{k=1}^{n-1}r_k=(-1)^n$, and that $\displaystyle\sum_{k=2}^nr_k=-2r_1$. – Lucian Dec 24 '13 at 16:41

First, we may note that if $(r_1,\ldots,r_n)$ is a solution for the degree $n$ problem, then $(r_1,\ldots,r_n,0)$ is a solution for the degree $n+1$ problem, and vice versa.

Second, as has already been pointed out by benh, the solution space for $r=(r_1,\ldots,r_n)$ is given by the solutions of the set of the $n$ polynomial equations $$\sigma_k(-r)=\sigma_k(-r_1,\ldots,-r_n)=r_k$$ where $\sigma_k$ is the $k$th symmetric function, and the left-hand side are the coefficients of $\prod_{i=1}^n(t-r_i)$. These have degrees $1,2,\ldots,n$.

In the generic case, you can count the number of solutions by multiplying the degrees of the polynomials, which tells us there should be $n!$ complex solutions. This assumes we count roots with multiplicities, that there are no roots "at infinity", and that equations form a complete intersection (i.e. the solution space consists of a discrete set of points and no higher-dimensional sets).

If the solution space has positive dimension, there would be solutions "at infinity". By "at infinity", I mean we embed $n$-space into the projective $n$-space, and the new point are the ones at infinity. A point at infinity can be described by a "line" (i.e. complex plane) through the origin: i.e. by $(sx_1,\ldots,s_xn)$ as $s$ goes to infinity, where the $x_i$ are not all zero. So, I need to show that there are no solutions at infinity.

More formally, the projective $n$-space is given by the ratios $[x_0:x_1:\ldots:x_n]$ where not all $x_i$ are zero, which is the same as saying that the projective point corresponds to the line (complex plane) $(sx_0,\ldots,sx_n)$ (i.e. these all represent the same point in the projective space). The regular $n$-space is the subspace given by $[1:x_1:\ldots:x_n]$, while the points at infinity are those with $x_0=0$.

If there are solutions at infinity, then the leading terms (i.e. top degree terms) of the $n$ polynomials $\sigma_k(-r)-r_k$ must all be zero at that point. For $k>1$, the leading term is simply $\sigma_k(-r)$. If $\sigma_n(-x)=0$ for some $x=(x_1,\ldots,x_n)$, then one of the $x_i$ must be zero; if also $\sigma_{n-1}(-x)=0$, then another one must be zero; and so on down to $\sigma_2(-x)=0$, hence $n-1$ of the $x_i$ must be zero. Finally, we have the last equation, $x_1+\sigma_1(x)=2x_1+x_2+\cdots+x_n=0$ with $n-1$ of the $x_i$ equal to zero, which forces the last one to be zero as well. Thus, there are no solutions at infinity.

So, counting complex solutions with multiplicity yields $n!$ solutions.

However, I suspect it would be a lot tougher to identify the real solutions.

• So if I understand you correctly, you have a set of $n$ equations, each of degree $1, 2, \dots, n$, so each has independently $1, 2, \dots, n$ solutions. Thus, when you multiply all of the possibilities together in a combinatorics-esque style, you get $n!$? (Sorry, I'm not very knowledgable about this) – MCT Mar 23 '14 at 1:00
• In $n$ dimensional space, the zero set of a degree $p$ polynomial will be a $n-1$ dimensional "surface" of degree $p$ (if "degree" is defined in the right way). If you take $k$ polynomials of degrees $p_1,\ldots,p_k$, you'll get a $n-k$ dimensional "surface" of degree $p_1\cdots p_k$. This is true for the generic case (i.e. except special cases). In the projective $n$ space, if you count solutions with multiplicity, the only special case is when the solution has higher dimensionality than $n-k$. – Einar Rødland Mar 23 '14 at 1:51
• Okay, thank you. Assuming no solution comes within the next couple days covering the real solutions, I will award you the bounty. – MCT Mar 24 '14 at 0:04

The coefficient $a_k$ of the polynomial $f$ is (up to a sign) the $k$-th elementary symmetric polynomial in $n$ variables evaluated at the complex roots of $f$, that is:

$$\begin{eqnarray} r_1 &=& -e_1(r_1,...,r_n)&=&r_1+...+r_n \\ r_2 &=& e_2(r_1,...,r_n)&=&r_1r_2+r_1r_3...+r_{n-1}r_n \\ r_n &=& (-1)^ne_n(r_1,...,r_n)&=&r_1\cdot \cdot \cdot r_n \end{eqnarray}$$

So from an algebraic-geometric point of view, you are asking for the vanishing set of the set of polynomials $$\Big(-e_1(X_1,...,X_n)-X_1\;,\;\dots\;,\;(-1)^ne_n(X_1,...,X_n)-X_n \Big).$$

As the elementary symmetric polynomials generate the ring of all symmetric polynomials (hence the name), the set of all solutions should be a finite algebraic subset of $\Bbb C^n$.

From an analytical point of view, the solutions are determined by these $n$ algebraic equations.

• Right, I understand as much. So are you contending that there are no generalizations, not even for the number of real/complex roots? – MCT Dec 24 '13 at 17:02
• My answer was intended to point on some structure I would start with to analyze the problem. Basically, I am contending that one could give algebraic geometry a chance to solve this problem, because I doubt that closed forms for the coefficients will help us here. – benh Dec 24 '13 at 17:27

Alpha confirms your counts for degree $3$. The number of complex solutions look like the triangular numbers. That is surprising-I would expect the factorials, as that is the product of the degrees of the equations. Unfortunately, when I went fourth degree, Alpha started only giving the value of one variable, not all four. I did get $23$ solutions out of Alpha (clicking on more roots) for the fourth degree equation. Five of them were $r_1=1$ but the others were all distinct. There were two other real solutions for $r_1$, but it didn't show the rest were real.

• I used fourth degree for alpha as well, which is how I got 5 real 10 complex. wolframalpha.com/input/… – MCT Dec 24 '13 at 16:39
• In the window with complex solutions there is a button for More Solutions which got me to 18 after a few clicks. I suspect one of the real solutions (probably $0,0,0,0$) is a double root, which gets us to 24 – Ross Millikan Dec 24 '13 at 16:45
• Also, why do you say the complex solutions look like triangular numbers? Is $1, 2, 6, 24$ not the factorials? – MCT Dec 24 '13 at 17:08
• @user92774: yes, but you had $2,6,10$ in there at the time. I was just noticing that they were triangular. – Ross Millikan Dec 24 '13 at 17:16

Using this code to generate the equations in Mathematica

n = 7;
vars = {a, b, c, d, e, f, g};
eqs = Table[(-1)^i SymmetricPolynomial[i, vars] - vars[[i]], {i, n}];
eqs // InputForm


and Macaulay2 to compute dimensions and degrees, we get:

Macaulay2, version 1.5
warning: sample Factory finite field addition table file missing, factorization may fail: /usr/share/Macaulay2/Core/factory/gftable.31.2
with packages: ConwayPolynomials, Elimination, IntegralClosure, LLLBases,
PrimaryDecomposition, ReesAlgebra, TangentCone

i1 : R = QQ[a, b, c]

o1 = R

o1 : PolynomialRing

i2 : I = ideal (-2*a - b - c, -b + a*b + a*c + b*c, -c - a*b*c);

o2 : Ideal of R

i3 : dim I

o3 = 0

i4 : degree I

o4 = 6

i5 : R = QQ[a, b, c, d];

i6 : I = ideal (-2*a - b - c - d, -b + a*b + a*c + b*c + a*d + b*d + c*d,
-c - a*b*c - a*b*d - a*c*d - b*c*d, -d + a*b*c*d);

o6 : Ideal of R

i7 : dim I

o7 = 0

i8 : degree I

o8 = 24

i9 : R = QQ[a, b, c, d, e];

i10 : I = ideal (-2*a - b - c - d - e, -b + a*b + a*c + b*c + a*d + b*d + c*d +
a*e + b*e + c*e + d*e, -c - a*b*c - a*b*d - a*c*d - b*c*d -
a*b*e - a*c*e - b*c*e - a*d*e - b*d*e - c*d*e,
-d + a*b*c*d + a*b*c*e + a*b*d*e + a*c*d*e + b*c*d*e,
-e - a*b*c*d*e);

o10 : Ideal of R

i11 : dim I

o11 = 0

i12 : degree I

o12 = 120

i13 : R = QQ[a, b, c, d, e, f];

i14 : I = ideal (-2*a - b - c - d - e - f, -b + a*b + a*c + b*c + a*d + b*d +
c*d + a*e + b*e + c*e + d*e + a*f + b*f + c*f + d*f + e*f,
-c - a*b*c - a*b*d - a*c*d - b*c*d - a*b*e - a*c*e - b*c*e -
a*d*e - b*d*e - c*d*e - a*b*f - a*c*f - b*c*f - a*d*f - b*d*f -
c*d*f - a*e*f - b*e*f - c*e*f - d*e*f,
-d + a*b*c*d + a*b*c*e + a*b*d*e + a*c*d*e + b*c*d*e + a*b*c*f +
a*b*d*f + a*c*d*f + b*c*d*f + a*b*e*f + a*c*e*f + b*c*e*f +
a*d*e*f + b*d*e*f + c*d*e*f, -e - a*b*c*d*e - a*b*c*d*f -
a*b*c*e*f - a*b*d*e*f - a*c*d*e*f - b*c*d*e*f, -f + a*b*c*d*e*f);

o14 : Ideal of R

i15 : dim I

o15 = 0

i16 : degree I

o16 = 720

i17 : R = QQ[a, b, c, d, e, f, g];

i18 : I = ideal (-2*a - b - c - d - e - f - g, -b + a*b + a*c + b*c + a*d + b*d +
c*d + a*e + b*e + c*e + d*e + a*f + b*f + c*f + d*f + e*f +
a*g + b*g + c*g + d*g + e*g + f*g, -c - a*b*c - a*b*d - a*c*d -
b*c*d - a*b*e - a*c*e - b*c*e - a*d*e - b*d*e - c*d*e - a*b*f -
a*c*f - b*c*f - a*d*f - b*d*f - c*d*f - a*e*f - b*e*f - c*e*f -
d*e*f - a*b*g - a*c*g - b*c*g - a*d*g - b*d*g - c*d*g - a*e*g -
b*e*g - c*e*g - d*e*g - a*f*g - b*f*g - c*f*g - d*f*g - e*f*g,
-d + a*b*c*d + a*b*c*e + a*b*d*e + a*c*d*e + b*c*d*e + a*b*c*f +
a*b*d*f + a*c*d*f + b*c*d*f + a*b*e*f + a*c*e*f + b*c*e*f +
a*d*e*f + b*d*e*f + c*d*e*f + a*b*c*g + a*b*d*g + a*c*d*g +
b*c*d*g + a*b*e*g + a*c*e*g + b*c*e*g + a*d*e*g + b*d*e*g +
c*d*e*g + a*b*f*g + a*c*f*g + b*c*f*g + a*d*f*g + b*d*f*g +
c*d*f*g + a*e*f*g + b*e*f*g + c*e*f*g + d*e*f*g,
-e - a*b*c*d*e - a*b*c*d*f - a*b*c*e*f - a*b*d*e*f - a*c*d*e*f -
b*c*d*e*f - a*b*c*d*g - a*b*c*e*g - a*b*d*e*g - a*c*d*e*g -
b*c*d*e*g - a*b*c*f*g - a*b*d*f*g - a*c*d*f*g - b*c*d*f*g -
a*b*e*f*g - a*c*e*f*g - b*c*e*f*g - a*d*e*f*g - b*d*e*f*g -
c*d*e*f*g, -f + a*b*c*d*e*f + a*b*c*d*e*g + a*b*c*d*f*g +
a*b*c*e*f*g + a*b*d*e*f*g + a*c*d*e*f*g + b*c*d*e*f*g,
-g - a*b*c*d*e*f*g);

o18 : Ideal of R

i19 : dim I

o19 = 0

i20 : degree I

o20 = 5040

• Ah. I have a sign wrong! I'll fix this later. – Mariano Suárez-Álvarez Dec 24 '13 at 17:34
• It looks like it is confirming that there are $n!$ solutions for degree $n$, at least up to $n=8$ Correct? – Ross Millikan Dec 24 '13 at 17:52