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.