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

(...)

The set of elements obtainable from $a_o, a_1, ... , a_{n-1}$ by $+, -,*,÷$ is the field $Q(a_0, a_1, ... , a_{n-1})$.

(...)

The goal of solution by radicals is then to extend $Q(a_0, a_1, ... , a_{n-1})$ by adjoining radicals until a field containing the roots $x_1, ... , x_n$ (roots of (*)) is obtained.


For example, the roots $x_1, x_2$ of the quadratic equation lie in the extension of $Q(a_0, a_1)=Q(x_1x_2, x_1+x_2)$ by the radical

$$\sqrt{a_1^2-4a_0}= \sqrt{(x_1+x_2)^2-4x_1x_2}=\sqrt{(x_1-x_2)^2}=±(x_1-x_2)$$


I have two questions concerning this last part (the one in between the lines).

  1. Where does the equality come from? I certainly would understand it if it wasn't for the most-left hand side.

  2. More importantly. What is the meaning of "$Q(a_0, a_1)=Q(x_1x_2, x_1+x_2)$"? By definition I can understand what the left hand side means, but I don't get the right hand side, and because of this, I think, the equality makes no sense to me.


It might be because I'm fairly new to the concept of Fields. Although I was able to get a grasp at what a group extension is. My knowledge in the subject goes as far as something like this: being $Q$ the group of the rational numbers, $Q[\sqrt{2}]$ would be an extension of $Q$ in which each the group would contain the roots of the polynomial $x^2-2$, and in which each element of the group could be written as $a+b\sqrt{2}$, where $a$ and $b$ are rational. Does it work in a similar way when we talkk about fields? And still, what would the equality "$Q(a_0, a_1)=Q(x_1x_2, x_1+x_2)$" mean?


I would really appreciate any help/ideas.

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  • $\begingroup$ I'm guessing somewhere in the text is that $a_n=1$? $\endgroup$ – Thomas Andrews Sep 8 '17 at 13:55
  • $\begingroup$ @Thomas Andrews. Yes. I'm sorry I didn't specified. $\endgroup$ – Leo Sep 8 '17 at 14:58
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If $S$ is a subset of $\mathbb{C}$, $\mathbb{Q}(S)$ denotes the smallest subfield of $\mathbb{C}$ containing $\mathbb{Q}$ and $S$.

One may show (that is a good exercise) that $\mathbb{Q}(S)$ is the set of all rational expressions in the elements of $S$ with coefficients in $\mathbb{Q}$.

When $S=\{z_1,\ldots,z_m\}$ is finite, this boils down to the set of elements of the form $$\frac{P(z_1,\ldots,z_m)}{Q(z_1,\ldots,z_m)},$$ where $P,Q\in \mathbb{Q}[T_1,\ldots,T_m]$, where $Q$ is such that $Q(z_1,\ldots,z_m)\neq 0.$

On the other hand, $\mathbb{Q}[S]$ denotes the smallest subringof $\mathbb{C}$ containing $\mathbb{Q}$ and $S$.

One may show (that is a good exercise also) that $\mathbb{Q}[S]$ is the set of all polynomial expressions in the elements of $S$ with coefficients in $\mathbb{Q}$.

When $S=\{z_1,\ldots,z_m\}$ is finite, this boils down to the set of elements of the form $$P(z_1,\ldots,z_m),$$ where $P\in \mathbb{Q}[T_1,\ldots,T_m]$.

For a single element $\alpha\in\mathbb{C}$, one may show that $\mathbb{Q}[\alpha]=\mathbb{Q}(\alpha)$ if and only if $\alpha$ is algebraic over $\mathbb{Q}$.

I strongly recommend you to learn the basics of field theory if you want to go further (I can imagine you want to understand the result of Abel-Galois on solvability of polynomial equations by radicals)

The equality $\mathbb{Q}(a_0,a_1)=\mathbb{Q}(x_1x_2,x_1+x_2)$ simply comes from the equalities $x_1+x_2=-a_1$ and $x_1x_2=a_0$ (expand $(T-x_1)(T-x_2)$ and compare coefficients)

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  • $\begingroup$ $a_1=-(x_1+x_2)$, but the minus sign doesn't complicate the equality. $\endgroup$ – Thomas Andrews Sep 8 '17 at 15:04
  • $\begingroup$ Yes, of course ! I've edited my message and removed the typo. $\endgroup$ – GreginGre Sep 9 '17 at 8:53

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