# Understanding a sentence in a Galois Theory paper about polynomials

I've been trying to understand a sentence of a math for some time. I'll be straightforward:

(*) $$x^n + a_{n-1}x^{n-1} +...+a_1x + a_0 = 0$$

(...)

If we denote the roots of (*) by $x_1, x_2, ... , x_n$ so that

$$(x-x_1)(x-x_2)...(x-x_n) = x^n + a_{n-1}x^{n-1} +...+a_1x + a_0$$

Then $a_0, ... , a_{n-1}$ are polynomial functions of $x_1, x_2, ... , x_n$ called elementary symmetric functions:

$$a_0 = (-1)^nx_1x_2...x_n , a_{n-1} = -(x_1 + x_2 + ... + x_n)$$

This last statement between the two lines is the one I have some issues with.

1. When the author claims that $a_0, ... , a_n$ are functions of $x_1, ... , x_n$, does he simply mean it in a way that $x_1, ... , x_n$ define what the elements $a_0, ... , a_n$ are?

2. And more importantly, where does this last equation come from? The only way I can think of making a_0 a function of $x_1, ... , x_n$ and $a_1, ... , a_n$ is to use (*) to write

$$a_0 = (-1)[x^n + a_{n-1}x^{n-1} +...+a_1x]$$

, but I can't seem to relate this last equality to the one the author wrote.

I would truly appreciate any help/thoughts!

• In case anyone wants to read the paper for some context (see page 1 & 2): math.jhu.edu/~smahanta/Teaching/Spring10/Stillwell.pdf – Leo Sep 8 '17 at 11:16
• 1. He means that there exists polynomials $P_0, P_1, \ldots, P_n$ in $n$ variables each (independent of the $a_i$ and of the $x_i$) such that each $i \in \left\{0,1,\ldots,n\right\}$ satisfies $a_i = P_i\left(x_1, x_2, \ldots, x_n\right)$. – darij grinberg Sep 8 '17 at 11:18
• The author is simply mentioning Vieta's relations. – Bernard Sep 8 '17 at 11:21
• Also, the field generated by the coefficients is a subfield of the field generated by the roots. And the field generated by the roots is in the algebraic closure of the field generated by the coefficients (this is the starting point of Galois theory). The same holds when the roots are seen as free variables. – reuns Sep 8 '17 at 11:30
• Endorsing @Bernard comment, this is Vieta's formulas en.wikipedia.org/wiki/Vieta%27s_formulas ... – rtybase Sep 8 '17 at 11:47

## 2 Answers

I assume that the author means: Given the roots $x_1,...,x_n$, I can calculate the $a_0,...,a_{n-1}$. What you are basically doing is just multiplicating out the right hand side of the equation: $$x^n + a_{n-1}x^{n-1} +...+a_1x + a_0=(x-x_1)(x-x_2)...(x-x_n)$$ Then the only term, which has degree $0$ on the right side is $\prod\limits_{i=1}^{n} (-x_i)=(-1)^n\prod\limits_{i=1}^{n}x_i$, thus it has to be equal to the term of degree $0$ on the left side, which is $a_0$. The only terms of degree $n-1$ on the right side are of the form $x^{n-1}\cdot (-x_i)$. Summing them up you get $(-\sum\limits_{i=1}^{n}x_i)x^{n-1}$ and comparing to the term of degree $n-1$ on the left side, $a_{n-1} x^{n-1}$, you get $a_{n-1}=-\sum\limits_{i=1}^{n}x_i$

For $1$, what darij said.

Just expand $(x-x_1)(x-x_2)\cdots (x-x_n)$. For small $n$, we have:

$$(x-x_1)(x-x_2) = x^2 - (x_1 + x_2) x + x_1 x_2 \\(x-x_1)(x-x_2)(x-x_3) = x^3 - (x_1 + x_2 + x_3)x^2 + (x_1 x_2 + x_1 x_3 + x_2 x_3)x - x_1 x_2 x_3$$

and the pattern reveals itself as the author claims:

$$(x-x_1)\cdots(x-x_n) = x^n - (x_1 + \cdots + x_n) x^{n-1} + \cdots + (-1)^n x_1 \cdots x_n$$

now compare these two expressions of $x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0$.