## 305 -Extensions by radicals (3)

March 8, 2009

3. Examples

Last lecture we defined what it means that a polynomial with coefficients in a field ${{\mathbb F}}$ is solvable by radicals over ${{\mathbb F}}$. Namely, there is a tower of extensions

$\displaystyle {\mathbb F}(t_1,\dots,t_k):{\mathbb F}(t_1,\dots,t_{k-1}):\dots:{\mathbb F}(t_1):{\mathbb F}$

where for each ${j,}$ ${1\le j\le k,}$ there is a positive integer ${m_j}$ such that ${t_j^{m_j}\in{\mathbb F}(t_1,\dots,t_{j-1}),}$ and such that ${{\mathbb F}^{p(x)}\subseteq{\mathbb F}(t_1,\dots,t_k).}$

## 305 -Solving cubic and quartic polynomials (2)

February 3, 2009

Let’s return to the problem of solving quartic polynomials. In the first lecture on this topic, we reduced the problem of solving an equation like

$x^4+ax^3+bx^2+cx+d=0$

to solving the similar problem

$y^4+py^2+qy+r=0,$

in which the coefficient of $y^3$ is zero. This is achieved by a simple translation, simply set $x=y-a/4$. This was motivated and explained in that lecture. Let us now see how we can approach this problem.

## 305 -2. Solving cubic and quartic polynomials.

January 28, 2009

1. We all know how to solve a linear equation such as $ax+b=0$, namely $x=-b/a$ (assuming $a\ne0$; if $a=0$ then either $b=0$ and any $x$ is a solution, or else there are no solutions). This was known to Babylonian and Persian mathematicians (with the usual caveats about the signs of $a,b$, since the notion of negative numbers had not been introduced yet.)

This is trivial, but there is a subtle point here:

• Some equations have no solutions.

If we are interested in solving polynomial equations in general, at some point we will need an argument justifying that we can. For now, let us proceed formally, assuming that we will always find solutions.

Just as with linear equations, we all know as well how to solve quadratics, such as $ax^2+bx+c=0$. Namely, we can factor $a$ out (if  $a=0$ we are in the linear case, so let’s assume that this is not the case) and then complete the square. We get $ax^2+bx+c=$

$\displaystyle a\left(x^2+\frac ba x+\frac ca\right)= a\left[\left(x+\frac b{2a}\right)^2+\left(\frac ca -\frac {b^2}{4a^2}\right)\right],$

so $ax^2+bx+c=0$ iff $(x+b/2a)^2=(b^2/4a^2)-(c/a)$, or

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},$

Another small subtlety appears here, namely, there is some inherent ambiguity in the meaning of the expression $\sqrt r$. We usually resolve this by “choosing a sign” of the square root. As long as we are looking at quadratic polynomials with integer (or rational, or real, or even complex) coefficients, there is a standard way of making this choice. In more general situations (in arbitrary fields) there is no such standard procedure.