There are many occasions where we can check whether a (set of) polynomial(s) $f_i$ satisfies certain properties, simply by evaluating a fixed polynomial on the coefficients of the $f_i$. Many times, the proofs of these results are based on deep theorems from algebraic geometry and other areas, that is, they are in no way trivial. My question is for you to please:

$\quad \quad$ Question 1: Provide examples like these with (possibly) their proofs.

Also, I would like to know whether

$\quad \quad$ Question 2: Is there a more uniform way to view, understand, or expect, these phenomena, in terms of Algebraic Geometry (or something else entirely) ?

We can start the list with the following theorems:


  1. A polynomial $$f=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0$$ has multiple roots if and only if the $n^{\text{th}}$-degree Discriminant $\Delta_n$ is $0$ for the coefficients $a_i$.

For instance for $f=ax^2+bx+c$, we have $\Delta_2=b^2-4ac$.

Proof: The discriminant $\Delta_n$ is a polynomial on the coefficients $a_i$. This is a special case of the next part as $\Delta(f)$ is really the resultant Res$(f,f')$.


  1. A set of $n$ polynomials $\{f_1,\cdots,f_n\}$, in variables $x_1,\cdots,x_n$, of given degrees $(d_1,\cdots,d_n)$ have a common root if and only if their Resultant is $0$: $$\operatorname{Res}(f_1,\cdots,f_n)=0$$

Proof: The fact that the resultant is a polynomial of the coefficients of the $f_i$'s can be proven via elementary methods but a more enlightening path relates it to the "Main result of Elimination Theory":

Theorem :(Projective Extension Theorem) Given a variety $W\subset \mathbb{C}^M\times\mathbb{P}^n$ and the projection map $\pi: \mathbb{C}^M\times\mathbb{P}^n\rightarrow \mathbb{C}^M$, the image $\pi (W)$ is a variety in $\mathbb{C}^M$.

Here, the idea is for $\mathbb{C}^M$ to represent the space of all sets of polynomials $\{f_i\}$ (via their coefficients) and $\mathbb{P}^n$ the values for the $x_i$ (after homogenizing maybe). Then, define $W$ as the pairs $\big( \{f_i\},{\bf x}\big)$ such that $f_i({\bf x})=0\ \forall i$. It is easy to see that $W$ is indeed a variety (its equation is just the (polynomial) evaluation of the $f_i$'s at ${\bf x}$). The theorem states that the image $\pi(W)$ (which corresponds to $f_i$'s having at least one non-trivial common root) is again a variety; that is, it is cut by a polynomial equation in $\mathbb{C}^M$, the space of polynomials, parametrised by their coefficients.


  1. A polynomial $f$ on $n$ variables $x_i$ and of degree $r$ is reducible, that is can be written as a product $$f(x_1,\cdots,x_n)=g(x_1,\cdots,x_n)\cdot h(x_1,\cdots,x_n)$$ of polynomials $h$ and $g$ of smaller degree.

Proof : I am not aware of any name for the resulting polynomial(s) on the coefficients of $f$. The proof uses the following corollary to the Projective extension theorem:

Corollary :[Eisenbud, Cor. 14.2] The image of a projective variety under a morphism is closed; more precisely if $Y$ is a projective variety over a field $k$, and $\pi: Y\rightarrow X$ is a $k$-morphism to a projective variety $X$, then $\pi (Y)$ is a closed subset of $X$ in the Zariski topology.

The idea is to consider the varieties $W_i$ that are formed by all (homogeneous) polynomials of degree $i$. Then, there is a map $W_i\times W_{n-i}\rightarrow W_n$ where we send $(f,g)$ to their product $f\cdot g$. This map is clearly a morphism (since multiplication of polynomials is a polynomial map on the coefficients). The image then, corresponds to polynomials that can be written as a product of a degree $i$ and a degree $n-i$ polynomials.

The aforementioned corollary implies then that the image is closed, hence given by polynomial equations (for details see section $\S$14.1 in Eisenbud's very beautiful Commutative Algebra with a view to a kill).


  1. A system of polynomials $(f_1,\cdots,f_n)$ is algebraically dependent, that is, there exists a polynomial $h\in \mathbb{C}[z_1,\cdots,z_n]$ such that $$h(f_1,\cdots,f_n)=0$$

Proof : The proof for this is the celebrated Jacobian criterion which states that the $f_i$ are algebraically dependent if and only if $$\operatorname{Jac}(f_1,\cdots,f_n)=0$$ where, of course, the jacobian polynomial being $0$ translates to certain polynomials on the coefficients of the $f_i$'s being $0$.

See for instance $\S$3.10 in Reflection Groups and Coxeter Groups by Humphreys, or the answer here, or the mathoverflow post here.


  1. A set of homogeneous polynomials $\big\{f_1({\bf x}),\cdots, f_n({\bf x})\big\}$, of given degrees $d_1,\cdots,d_n$, is

    • a homogeneous system of parameters (h.s.o.p.) for the ring $\mathbb{C}[x_1,\cdots,x_n]$,

    • or equivalently, the map $$\begin{alignat*}{1} \mathbb{C}^n&\rightarrow \mathbb{C}^n \\ {\bf x}:=(x_1\cdots,x_n) &\rightarrow \big(f_1({\bf x}),\cdots, f_n({\bf x})\big) \end{alignat*}$$ is a finite morphism,

    • or equivalently, the only preimage (for the map above) of the origin is the origin, i.e. $${\bf f}^{-1}({\bf 0})={\bf 0}$$

Proof : The proof here is more difficult. First notice that being a h.s.o.p. means that the algebra $\displaystyle \mathbb{C}[x_1,\cdots,x_n]/(f_1,\cdots,f_n)$ is finite dimensional (as a vector space).

Therefore, a tuple $(f_1,\cdots,f_n)$ forms a h.s.o.p. iff all but finitely many of the monomials $x_1^{a_1}\cdots x_n^{a_n}$ are in the ideal generated by the $f_i$'s. For that matter, it is enough for the ideal $(f_1,\cdots,f_n)$ to contain all the monomials of some degree $N$.

Now, this can be phrased as the existence of a solution to a linear system involving the coefficients of the $f_i$'s but we need to know a suitable $N$. To the rescue comes the following characterization for regular sequences of homogeneous polynomials (notice that in our setting, a maximal regular sequence of homogeneous polynomials and a h.s.o.p. mean the same thing since $\mathbb{C}[x_1,\cdots,x_n]$ is a regular (homogeneous-)local ring):

Theorem :[Stanley: Hilbert functions of graded algebras, Corol. 3.2] A tuple of homogeneous polynomials $(f_1,\cdots,f_n)$ is a regular sequence for the ring $\mathbb{C}[x_1,\cdots,x_n]$ if and only if the Hilbert series of the quotient is given by: $$\operatorname{Hilb}\big(\mathbb{C}[x_1,\cdots,x_n]/(f_1,\cdots,f_n),q \big)= \prod_{i=1}^n \frac{q^{d_i}-1}{q-1}$$ where $d_i:=\operatorname{deg}(f_i)$.

Therefore, the smallest $N$ for which $(f_1,\cdots,f_n)$ must contain all monomials of degree $N$ is $$N=\sum (d_i-1)+1$$

Remark : Actually, for a tuple of (not necessarily homogeneous) polynomials, to be a regular sequence, is also a polynomial property on the coefficients but I do not have a reference for a proof of that. (?)


  • $\begingroup$ For 1., you also need to fix the degree. The better statement would be something along the lines of "the degree-$n$ homogenization has a repeated factor". $\endgroup$ – darij grinberg Jun 4 '15 at 20:50
  • $\begingroup$ Without thinking much, I would suspect that an answer to your question 2 could go along the lines of "linear algebra + Noetherian-type bounds". By which I mean the following: Write the condition on the polynomials as a statement that a certain system of linear equations (a priori infinitely many, in infinitely many variables) has a nontrivial solution. Argue using some kind of bounds on degrees that this system can be reduced to a finite one. Then, rewrite it as the vanishing of a determinant. But this is a very vague idea, and does not trivialize the question! $\endgroup$ – darij grinberg Jun 4 '15 at 20:53
  • $\begingroup$ @darijgrinberg Yes, of course the degree should be fixed, I mention it at some points but I also skipped it occasionally since the post was getting too big. I agree that given a property that "feels" as if it can be phrased in terms of linear algebra and monomials, the rest should work. But a priori these things seem be too complicated to reduce down to simple conditions for monomials (h.s.o.p.'s, common solutions..)! I feel some of these results are pretty remarkable... $\endgroup$ – Theo Jun 4 '15 at 21:07

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