How can I use Fundamental Theorem of Symmetric Polynomials to factor polynomials?

Question

How can I use The fundamental theorem of symmetric polynomials (or its proof) to factor symmetric polynomials?

The link I've given to the theorem uses elaborate wordings using 'rings', 'isomorphic', etc.

I completely understand those objects or describings are needed to have a deep understanding, but could anyone try, if it is possible, to explain simply how I could use the theorem to, e.g., factor

$(a^4+b^4)(a^2+b^2)-(a^3+b^3)^2 = a^2b^2(a^2+b^2-2ab)$

without understanding what rings are? I only wish to be able to practically use it.

Answer 1

If I understand correctly, the fundamental theorem of symmetric polynomials says a symmetric polynomial of $x_1, x_2, \dots, x_n$ can be written as a polynomial of $$\quad e_1 = \sum_i x_i,\quad e_2 = \sum_{i < j}x_ix_j, \quad \dots, \quad e_n = \prod_i x_i.$$

For example, symmetric polynomial $(a^4+b^4)(a^2+b^2)-(a^3+b^3)^2$ can be written as a polynomial of $x = a + b, y = ab$. In this example, inductively one can get $a^n + b^n$ in terms of $x,y$. Note that $$a^2 + b^2 = (a+b)^2-2ab = x^2-2y,$$ $$a^3 + b^3 = (a^2 + b^2)(a+b) - ab(a+b) = (x^2-2y)x - yx = x^3-3xy,$$ $$a^4+b^4 = (a^3+b^3)(a+b)-ab(a^2+b^2) = (x^3-3xy)x-y(x^2-2y)=x^4-4x^2y+2y^2.$$

Therefore we have $$(a^4+b^4)(a^2+b^2)-(a^3+b^3)^2 = (x^2-2y)(x^4-4x^2y+2y^2)-(x^3-3xy)^2 = x^2y^2-4y^3.$$

In general, once a symmetric polynomial is written in terms of elementary symmetric polynomials, it might be easier to observe factors. In this example, $x^2y^2-4y^3 = y^2(x^2-4y)$.

Warning One might want to substitute the original variables back to see if certain factors can be factorized further. In the example, $x^2-4y$ is irreducible, however $x^2-4y = a^2+b^2-2ab = (a-b)^2$ is reducible.

Answer 2

Note: As already noted by @ZilinJ the Fundamental Theorem of Symmetric Polynomials guarantees the unique representation of symmetric polynomials $P(x_1,\ldots,x_n)$ as polynomial $Q$ in the elementary symmetric polynomials $e_1,e_2,\ldots,e_n$ in $n$ variables \begin{align*} P(x_1,x_2, \ldots,x_n)=Q(e_1,e_2,\ldots,e_n) \end{align*} with \begin{align*} e_1&=e_1(x_1,\ldots,x_n)=x_1+x_2+\cdots+x_n\\ e_2&=e_2(x_1,\ldots,x_n)=x_1x_2+x_1x_3+\cdots+x_{n-1}x_n\\ &\ldots\\ e_n&=e_n(x_1,\ldots,x_n)=x_1x_2\cdots x_n \end{align*}

$$$$

This answer introduces a method to systematically transform a symmetric polynomial into a polynomial representation by elementary symmetric polynomials. It's based on Paul Garrets algebra course section $15$ Symmetric polynomials.

Let's start with OPs example and then continue with a slightly more complex one in order to better see how the method works.

OPs symmetric polynomial $P(a,b)$ is

\begin{align*} P(a,b)&=(a^4+b^4)(a^2+b^2)-(a^3+b^3)^2\\ &=a^4b^2-2a^3b^3+a^2b^4\\ &=a^2b^2(a^2-2ab+b^2) \end{align*}

We consider the elementary symmetric polynomials in $2$ variables $a,b$: \begin{align*} e_1&=e_1(a,b)=a+b\\ e_2&=e_2(a,b)=ab \end{align*} We observe, that a factor of $P(a,b)$ is already given as symmetric polynomial \begin{align*} P(a,b)=e_2(a,b)^2\cdot(a^2-2ab+b^2) \end{align*} and we put the focus on \begin{align*} f(a,b)=a^2-2ab+b^2 \end{align*}

The method: If it's obvious to represent $f(a,b)$ via elementary symmetric polynomials we are finished. Otherwise simplify $f(a,b)$ by setting the variable $b=0$. This reduces the number of variables by one and wie obtain a function \begin{align*} q(a):=f(a,0)=a^2 \end{align*} Find a representation of the polynomial $q(a)$ as polynomial $Q(a)$ in elementary symmetric polynomials $e_1(a)=a$ \begin{align*} q(a)=Q(e_1(a))=e_1(a)^2=a^2 \end{align*} Consider the polynomial $g(a,b)$ with \begin{align*} g(a,b)=Q(e_1(a,b))=e_1(a,b)^2 \end{align*} Then \begin{align*} \frac{f(a,b)-g(a,b)}{e_2(a,b)}=\frac{a^2-2ab+b^2-(a+b)^2}{ab}=-4\tag{1} \end{align*} is a polynomial of lower total degree than $f$. Now apply the method on this polynomial with lower degree.

Since the polynomial $-4$ is already simple enough, we can calculate $f(a,b)$ as

\begin{align*} f(a,b)=-4e_2(a,b)+g(a,b)=-4e_2(a,b)+e_1(a,b)^2 \end{align*}

We finally obtain a representation of OPs polynomial $P(a,b)$ according to (1) as polynomial of elementary symmetric polynomials

\begin{align*} P(a,b)&=e_2(a,b)^2\cdot(-4e_2(a,b)+e_1(a,b)^2)\\ &=-4e_2(a,b)^3+e_2(a,b)\cdot e_1(a,b)^2\\ &=-4(ab)^3+(ab)^2(a+b)^2 \end{align*}

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Second example: $f_1(a,b,c)=a^3+b^3+c^3$

We show \begin{align*} f_1(a,b,c)&=3e_3(a,b,c)-3e_2(a,b,c)e_1(a,b,c)+e_1(a,b,c)^3\\ &=3(abc)-3(ab+ac+bc)(a+b+c)+(a+b+c)^3 \end{align*}

We calculate this slightly more complex example by applying the method in two steps.

We consider the elementary symmetric polynomials in $3$ variables $a,b,c$: \begin{align*} e_1&=e_1(a,b,c)=a+b+c\\ e_2&=e_2(a,b,c)=ab+ac+bc\\ e_3&=e_3(a,b,c)=abc\\ \end{align*}

The method:

Step $1$: It is not obvious to represent $f_1(a,b,c)$ via elementary symmetrical polynomials and so we apply the method by setting the variable $c=0$. This reduces the number of variables by one and we obtain a function \begin{align*} q_1(a,b):=f_1(a,b,0)=a^3+b^3 \end{align*}

Step $2$: According to the method we should find a representation of the polynomial $q_1(a,b)$ as polynomial $Q_1(a,b)$ in elementary symmetric polynomials $e_1(a,b)$ and $e_2(a,b)$.

Since we don't see a representation immediately, we consider $q_1(a,b)$ as polynomial $f_2(a,b):=q_1(a,b)$ and apply the first part of the method again.

$$$$

We start with \begin{align*} f_2(a,b)=a^3+b^3 \end{align*} We again reduce the number of variables by one and we obtain a function \begin{align*} q_2(a):=f_2(a,0)=a^3 \end{align*} Now it's easy to see, that $q_2(a)$ can be represented via the elementary symmetric polynomial $e_1(a)=a$ \begin{align*} q_2(a)=Q_2(e_1(a))=e_1(a)^3=a^3 \end{align*} We consider the polynomial $g_2(a,b)$ with \begin{align*} g_2(a,b)=Q_2(e_1(a,b))=e_1(a,b)^3=(a+b)^3 \end{align*} Then \begin{align*} \frac{f_2(a,b)-g_2(a,b)}{e_2(a,b)}=\frac{a^3+b^3-(a+b)^3}{ab}=-3(a+b)=-3e_1(a,b)\tag{2} \end{align*} is a polynomial of lower total degree than $f_2$.

We observe according to (2)

\begin{align*} f_2(a,b)=-3e_1(a,b)e_2(a,b)+g_2(a,b)=-3e_1(a,b)e_2(a,b)+e_1(a,b)^3\tag{3} \end{align*} We have systematically found a representation of $q_1(a,b)=f_2(a,b)$ as polynomial of elementary symmetric polynomials and go on with step 1.

Step $1$ continued:

Since $q_1(a,b)=a^3+b^3=f_2(a,b)$ we obtain according to (3) \begin{align*} q_1(a,b)=-3e_1(a,b)e_2(a,b)+e_1(a,b)^3 \end{align*} we go on with defining \begin{align*} g_1(a,b,c)&=-3e_2(a,b,c)e_1(a,b,c)+e_1(a,b,c)^3\\ &=-3(ab+ac+bc)(a+b+c)+(a+b+c)^3\\ &=a^3+b^3+c^3-3abc \end{align*}

We calculate \begin{align*} \frac{f_1(a,b,c)-g_1(a,b,c)}{e_3(a,b,c)}&=\frac{f_1(a,b,c)+3e_2(a,b,c)e_1(a,b,c)-e_1(a,b,c)^3}{abc}\\ &=\frac{(a^3+b^3+c^3)-(a^3+b^3+c^3-3abc_)}{abc}\\ &=3\\ \end{align*} and we finally conclude \begin{align*} f_1(a,b,c)&=3e_3(a,b,c)+g_1(a,b,c)\\ &=3e_3(a,b,c)-3e_2(a,b,c)e_1(a,b,c)+e_1(a,b,c)^3\\ &=3(abc)-3(ab+ac+bc)(a+b+c)+(a+b+c)^3 \end{align*}

Answer 3

I will give you another algoritm to find the factorization. I'll use your polynomial to show the way it works. For a polynomial in $k$ variables the algorithm can be described as:

1. Order the terms of your polynomial in lexicographic ordening.
2. Suppose the highest term is $X_{1}^{n_1}\cdots X_{k}^{n_k}$, then subtract $e_{1}^{n_1-n_2}e_{2}^{n_2-n_3}\cdots e_{k}^{n_k}$.
3. Remember this term and head back to step 1 with your new polynomial. If the new polynomial is zero, you are finished.
4. Add all the terms found to get your representation for the original polynomial.

I won't prove here that it works, I will just show you how to apply it. The lexicographic ordening means that we start with the term which has highest $n_1$, then if two have the same value for $n_1$ we choose the one with highest $n_2$ etc.

Your polynomial: $f = (a^4+b^4)(a^2+b^2)−(a^3+b^3)^2 = a^4b^2 -2a^3b^3+ a^2b^4$, put in the lexicographic ordening. Next we create \begin{equation}f_1 = f - e_{1}^{4-2}e_{2}^{2} = f-(a+b)^2(ab)^2 = (a^4b^2-2a^3b^3+ a^2b^4) - (a^4b^2+2a^3b^3 + a^2b^4) = 0. \end{equation} Since this is zero we are done after one step and conclude that $f = f_1 + e_{1}^{4-2}e_{2}^{2} = 0 + e_{1}^{4-2}e_{2}^{2} = (a+b)^2(ab)^2$.

Answer 4

I just wanna add an easy way for readers: (it can be used in Projective Spaces in Algebraic Geometry)

First put $b=1$ (this is dehomogenize). Then $$(a^4+1)(a^2+1)-(a^3+1)^2=a^2(a^2+1-2a).$$ Now homoginize it to degree $6$: $$a^2b^2(a^2+b^2-2ab).$$