# proof of fundamental theorem of symmetric polynomials

Let $P:=P({x}_{1},{x}_{2},\mathrm{\dots},{x}_{n})$ be an arbitrary symmetric polynomial^{} in ${x}_{1},{x}_{2},\mathrm{\dots},{x}_{n}$. We can assume that $P$ is homogeneous^{} (http://planetmath.org/Polynomial^{}), because if $P={P}_{1}+{P}_{2}+\mathrm{\dots}+{P}_{m}$ where each ${P}_{i}$ is homogeneous and if the theorem (http://planetmath.org/FundamentalTheoremOfSymmetricPolynomials) is true for each ${P}_{i}$, it is evidently true for the sum $P$, too.

Let the degree (http://planetmath.org/Polynomial) of $P$ be $d$. For any two terms

$$M:={c}_{\mu}{x}_{1}^{{\mu}_{1}}{x}_{2}^{{\mu}_{2}}\mathrm{\cdots}{x}_{n}^{{\mu}_{n}},N:={c}_{\nu}{x}_{1}^{{\nu}_{1}}{x}_{2}^{{\nu}_{2}}\mathrm{\cdots}{x}_{n}^{{\nu}_{n}}$$ |

of $P$, if the first of the differences^{}

$${\mu}_{1}-{\nu}_{1},{\mu}_{2}-{\nu}_{2},\mathrm{\dots},{\mu}_{n}-{\nu}_{n},$$ |

which differs from 0, is positive, we say that $M$ is *higher* than $N$. Since, of cource, the terms of $P$ have been merged, always one of two arbitrary terms is higher than the other. The higherness is obviously transitive^{} (http://planetmath.org/Transitive3). Thus there is a certain *highest* term

$$A:={c}_{\alpha}{x}_{1}^{{\alpha}_{1}}{x}_{2}^{{\alpha}_{2}}\mathrm{\cdots}{x}_{n}^{{\alpha}_{n}}$$ |

in $P$. Then we have

$${\alpha}_{1}+{\alpha}_{2}+\mathrm{\dots}+{\alpha}_{n}=d,$$ |

$${\alpha}_{1}\ge {\alpha}_{2}\ge \mathrm{\dots}\ge {\alpha}_{n}.$$ |

In fact, if e.g. ${\alpha}_{2}>{\alpha}_{1}$, then the term

$${c}_{\alpha}{x}_{2}^{{\alpha}_{1}}{x}_{1}^{{\alpha}_{2}}{x}_{3}^{{\alpha}_{3}}\mathrm{\cdots}{x}_{n}^{{\alpha}_{n}}={c}_{\alpha}{x}_{1}^{{\alpha}_{2}}{x}_{2}^{{\alpha}_{1}}{x}_{3}^{{\alpha}_{3}}\mathrm{\cdots}{x}_{n}^{{\alpha}_{n}},$$ |

which is obtained from $A$ by changing ${x}_{1}$ and ${x}_{2}$ with each other, would be higher than $A$.

For proving the fundamental theorem (http://planetmath.org/FundamentalTheoremOfSymmetricPolynomials), we form now the homogeneous polynomial^{}

$${Q}_{\alpha}:={c}_{\alpha}{p}_{1}^{{i}_{1}}{p}_{2}^{{i}_{2}}\mathrm{\cdots}{p}_{n}^{{i}_{n}}$$ |

and we will show that the exponents (http://planetmath.org/Exponentiation^{}) ${i}_{j}$ can be determined such that the highest term in ${Q}_{\alpha}$ is same as in $P$.

It is easily seen that the highest term of a product^{} of homogeneous symmetric polynomials is equal to the product of the highest terms of the factors. Since the highest term of

$${p}_{1}\mathit{\hspace{1em}}\text{is}\mathit{\hspace{1em}}{x}_{1},$$ |

$${p}_{2}\mathit{\hspace{1em}}\text{is}\mathit{\hspace{1em}}{x}_{1}{x}_{2},$$ |

$$\mathrm{\cdots}\mathit{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{\cdots}$$ |

$${p}_{n}\mathit{\hspace{1em}}\text{is}\mathit{\hspace{1em}}{x}_{1}{x}_{2}\mathrm{\cdots}{x}_{n},$$ |

therefore the highest term of

$${p}_{1}^{{i}_{1}}\mathit{\hspace{1em}}\text{is}\mathit{\hspace{1em}}{x}_{1}^{{i}_{1}},$$ |

$${p}_{2}^{{i}_{2}}\mathit{\hspace{1em}}\text{is}\mathit{\hspace{1em}}{x}_{1}^{{i}_{2}}{x}_{2}^{{i}_{2}},$$ |

$$\mathrm{\cdots}\mathit{\hspace{1em}\hspace{1em}\hspace{1em}}\mathrm{\cdots}$$ |

$${p}_{n}^{{i}_{n}}\mathit{\hspace{1em}}\text{is}\mathit{\hspace{1em}}{x}_{1}^{{i}_{n}}{x}_{2}^{{i}_{n}}\mathrm{\cdots}{x}_{n}^{{i}_{n}}$$ |

and thus the highest term of ${Q}_{\alpha}$ is

$${c}_{\alpha}{x}_{1}^{{i}_{1}+{i}_{2}+\mathrm{\cdots}+{i}_{n}}{x}_{2}^{{i}_{2}+\mathrm{\cdots}+{i}_{n}}\mathrm{\cdots}{x}_{n}^{{i}_{n}}.$$ |

This term coincides with the highest term of $P$, when one determines the numbers ${i}_{j}$ from the equations

$\{\begin{array}{cc}{i}_{1}+{i}_{2}+\mathrm{\dots}+{i}_{n}={\alpha}_{1}\hfill & \\ {i}_{2}+\mathrm{\dots}+{i}_{n}={\alpha}_{2}\hfill & \\ \mathrm{\cdots}\hspace{1em}\hspace{1em}\mathrm{\cdots}\hfill & \\ {i}_{n}={\alpha}_{n}.\hfill & \end{array}$ |

Subtracting here the second equation from the first, the third equation from the second and so on, the result is

$${i}_{1}={\alpha}_{1}-{\alpha}_{2},{i}_{2}={\alpha}_{2}-{\alpha}_{3},\mathrm{\dots},{i}_{n-1}={\alpha}_{n-1}-{\alpha}_{n},{i}_{n}={\alpha}_{n},$$ |

which are nonnegative integers. Hence we get the homogeneous symmetric polynomial

$${Q}_{\alpha}={c}_{\alpha}{p}_{1}^{{\alpha}_{1}-{\alpha}_{2}}{p}_{2}^{{\alpha}_{2}-{\alpha}_{3}}\mathrm{\cdots}{p}_{n-1}^{{\alpha}_{n-1}-{\alpha}_{n}}{p}_{n}^{{\alpha}_{n}}$$ |

having the same highest term as $P$, and consequently the difference

$$P-{Q}_{\alpha}:={P}_{\alpha}$$ |

is a homogeneous symmetric polynomial of degree $d$ having the highest term lower than in $P$. If then

$${c}_{\beta}{x}_{1}^{{\beta}_{1}}{x}_{2}^{{\beta}_{2}}\mathrm{\cdots}{x}_{n}^{{\beta}_{n}}$$ |

is the highest term of ${P}_{\alpha}$ and one denotes

$${Q}_{\beta}:={c}_{\beta}{p}_{1}^{{\beta}_{1}-{\beta}_{2}}{p}_{2}^{{\beta}_{2}-{\beta}_{3}}\mathrm{\cdots}{p}_{n-1}^{{\beta}_{n-1}-{\beta}_{n}}{p}_{n}^{{\beta}_{n}},$$ |

one infers as above that the difference

$${P}_{\alpha}-{Q}_{\beta}:={P}_{\beta}$$ |

is a homogeneous symmetric polynomial of degree $d$ having the highest term lower than in ${P}_{\alpha}$. Continuing similarly, one finally (after a finite amount of steps) shall come to a difference which is equal to 0. Accordingly one obtains

$$P={Q}_{\alpha}+{Q}_{\beta}+\mathrm{\dots}+{Q}_{\omega}:=Q({p}_{1},{p}_{2},\mathrm{\dots},{p}_{n}).$$ |

The degree of ${Q}_{\alpha}$ with respect to the elementary symmetric polynomials is

$$({\alpha}_{1}-{\alpha}_{2})+({\alpha}_{2}-{\alpha}_{3})+\mathrm{\dots}+({\alpha}_{n-1}-{\alpha}_{n})+{\alpha}_{n}={\alpha}_{1}.$$ |

Similarly, the degree of ${Q}_{\beta}$ is ${\beta}_{1}$ which is $\le {\alpha}_{1}$; thus one infers that the degree of $Q$ is equal to ${\alpha}_{1}$. This number is also the degree of the highest term of $P$ and as well the degree of $P$ itself, with respect to ${x}_{1}$.

The preceding construction implies immediately that the coefficients of $G$ are elements of the ring determined by the coefficients of $P$. We have still to prove the uniqueness of $Q$. Let’s make the antithesis that $P$ may be represented also by another polynomial in ${p}_{1},{p}_{2},\mathrm{\dots},{p}_{n}$ which differs from $Q$. Forming the difference of it and $Q$ we get an equation of the form

$$0=\sum _{i}{g}_{i}{p}_{1}^{{i}_{1}}{p}_{2}^{{i}_{2}}\mathrm{\cdots}{p}_{n}^{{i}_{n}},$$ |

where the coefficients are distinct from zero. The equation becomes identical if one expresses ${p}_{1},{p}_{2},\mathrm{\dots},{p}_{n}$ in it via the indeterminates ${x}_{1},{x}_{2},\mathrm{\dots},{x}_{n}$. The general term of the right hand side of the equation is a homogeneous symmetric polynomial in those indeterminates; if its highest term is ${g}_{i}{p}_{1}^{{\lambda}_{1}}{p}_{2}^{{\lambda}_{2}}\mathrm{\cdots}{p}_{n}^{{\lambda}_{n}},$ one infers as before that

$${i}_{1}={\lambda}_{1}-{\lambda}_{2},{i}_{2}={\lambda}_{2}-{\lambda}_{3},\mathrm{\dots},{i}_{n-1}={\lambda}_{n-1}-{\lambda}_{n},{i}_{n}={\lambda}_{n}.$$ |

Thus, distinct addends of the sum cannot have equal highest terms. It means that the highest term of the sum appears only in one of the addends of the sum. This is, however, impossible, because after the substitution of ${x}_{i}$s the equation would not be identical. Consequently, the antithesis is wrong and the whole fundamental theorem of symmetric polynomials has been proved.

## References

- 1 K. Väisälä: Lukuteorian ja korkeamman algebran alkeet. Tiedekirjasto No. 17. Kustannusosakeyhtiö Otava, Helsinki (1950).

Title | proof of fundamental theorem of symmetric polynomials |
---|---|

Canonical name | ProofOfFundamentalTheoremOfSymmetricPolynomials |

Date of creation | 2016-02-22 13:29:31 |

Last modified on | 2016-02-22 13:29:31 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 14 |

Author | pahio (2872) |

Entry type | Proof |

Classification | msc 12F10 |

Classification | msc 13B25 |

Synonym | proof of fundamental theorem of symmetric functions |