A description for the ideal generated by the symmetric polynomials in two variables

This article is part of my migration effort, moving some of my articles over from the excellent Functor Network.

The goal of this short post is to convince myself that the ideal generated by $\operatorname{Sym}^+$ inside of $\mathbb{C}[x,y]$ can be more simply described as the ideal generated by $xy$ and $x+y$, i.e. $$\langle \operatorname{Sym}^+\rangle = \langle xy,x+y\rangle.$$

As a shorthand, set $A = \mathbb{C}[x,y]$. The ring $A$ is graded: $$A = \bigoplus_{d = 0}^\infty A^{(d)},$$ where $A^{(d)}$ is the $\mathbb{C}$-module consisting of the homogeneous polynomials of degree $d$: $$A^{(d)} = \mathbb{C}\{x^ay^b \mid a + b = d\}.$$

In general, a polynomial is said to be symmetric when it is invariant under any permutation of the variables. In our case, a polynomial $p(x,y) \in A$ is symmetric when $p(x,y) = p(y,x)$. For instance, $x^3 + y^3 + 2xy$ is symmetric while $x+y^2$ is not. The product and difference of two symmetric polynomials is also a symmetric polynomial. Also, $1$ is a trivial example of a symmetric polynomial. Hence the set of all symmetric polynomials is a subring of $A$, which we denote by $\operatorname{Sym}$. This subring is naturally graded: $$\operatorname{Sym} = \bigoplus_{d=0}^\infty \operatorname{Sym}^{(d)}$$ where $\operatorname{Sym}^{(d)} = \operatorname{Sym} \cap A^{(d)}$ is the set of symmetric homogeneous polynomials of degree $d$.

Now we restrict our attention to the set of symmetric polynomials which have $0$ as a root. This is exactly the set $$\operatorname{Sym}^+ = \bigoplus_{d=1}^\infty \operatorname{Sym}^{(d)}.$$

Let’s prove that every element in $\operatorname{Sym}^{+}$ can be written as an $A$-linear combination of $xy$ and $x+y$, which will show the first equation between generated ideals at the top of this post holds. Concretely, every element $p(x,y)$ in $\operatorname{Sym}^+$ can be written as some sum of homogeneous elements, all of degree at least one. If I can show that each of these homogeneous elements can be written in the form $$\text{[some polynomial]}\cdot (x+y) + \text{[some other polynomial]}\cdot xy,$$ then simply by grouping together terms in $x+y$ and terms in $xy$ and factoring out, I will obtain an expression for $p(x,y)$ as an $A$-linear combination of $x+y$ and $xy$. So we can reduce the problem to $p(x,y)$ being an homogeneous polynomial of degree $d \geq 1$. We can even do more. Recall that $\operatorname{Sym}^{(d)}$ is a vector space over $\mathbb{C}$, and if we can show every basis element can be written as a linear combination like we want, then we have shown $p(x,y)$ can be written like that as well. Hence we have reduced the problem to showing that for any $d \geq 1$, some basis of $\operatorname{Sym}^{(d)}$ can be written as an $A$-linear combination of $x+y$ and $xy$.

Let’s chose the $m$-basis. I hope to write some post about this basis. However I want to keep this one short, so here are the basics for future quick recalling. Let’s say that a partition (French: partage) of some natural number $n \geq 0$ is some list $\lambda = (\lambda_1, \dots, \lambda_k)$ of natural numbers with $\lambda_1 \geq \dots \geq \lambda_k \geq 0$ and such that $\lambda_1 + \dots + \lambda_k = n$. The integer $k$ is the numbers of parts of $\lambda$, which we may write as $\ell(\lambda)$. To indicate that $\lambda$ is a partition of $n$, we write $\lambda \vdash n$.

Fix some degree $d \geq 1$. To build a basis for $\operatorname{Sym}^{(d)}$, pick some $\lambda \vdash d$ with $\ell(\lambda) = 2$ (if needed, extend the partition with a second part of zero length). For instance, if $d=3$, then possible choices of partition are $(3,0)$ and $(2,1)$. Now write $$m_\lambda = x^{\lambda_1}y^{\lambda_2} + x^{\lambda_2}y^{\lambda_1}.$$ Collecting these polynomials for all possible partitions of $d$ gives you a basis for $\operatorname{Sym}^{(d)}$. Note that what I’ve just written is a special case of the more general construction using the Reynolds symmetrization operator when there are more than two variables. Anyways, in the $d=3$ example, the basis is given as $\{x^3+y^3, x^2y+xy^2\}$.

Let’s get back to our original problem. Recall: fixing a degree $d \geq 1$, we need to show that each $m_\lambda$ can be written in the form $p(x,y)(x+y) + r(x,y)xy$ for some polynomials $p$ and $q$ that obviously depend on $m_\lambda$. Here’s the argument, which is quite simple after all this yapping. Suppose $\lambda$ has two non-zero parts (eg. $\lambda=(2,1)$). Whatever $m_\lambda$ is, it is garanteed by construction that each monomial in $m_\lambda$ is divisible by both $x$ and $y$; hence we can factor $xy$ out of each monomial to obtain $$m_{(\lambda_1,\lambda_2)} = xy\cdot m_{(\lambda_1 - 1, \lambda_2 -1)}.$$ For instance, $$m_{(2,1)} = x^2y+xy^2 = xy(x+y) = xy\cdot m_{(1,0)}.$$ What if $\lambda$ has a single non-zero part? Then $m_\lambda = m_{(d)}$ looks like this: $$m_{(d)} = x^d + y^d.$$ Newton to the rescue: $$\begin{align*} x^d+y^d &= (x+y)^d - \sum_{i=1}^{d-1}{d\choose i}x^{d-i}y^i\\ &= (x+y)^d - xy\sum_{i=1}^{d-1}{d\choose i}x^{d-i-1}y^{i-1}. \end{align*}$$ So we win. The only subtelty here is, what if $d=1$ so then inside of the sum we have a $x^{-1}$ term? Well, if $d=1$ then the sum goes from $i=1$ to $i=0$, which means by convention that it’s the empty sum, which is zero. Anyways, the case $d=1$ can be done separately.