Fields and Quadratic Polynomials

2009 October 4

by Dr. Nichtgegeben

A field is an algebraic system very much like the usual systems of numbers you are most familiar with. To be slightly more precise, a field is a set with operations of $+$ and $\cdot$ such that both operations are associative and commutative, there is an element $0$ which acts like an additive identity, there is a number $1 \neq 0$ which acts like a multiplicative identity, each element has an additive inverse, each nonzero element has a multiplicative inverse, and, finally, the two operations are connected by the usual distributive laws. Three well-known examples are the rational numbers (denoted $\mathbb{Q}$ ), the real numbers (denoted $\mathbb{R}$ ), and the complex numbers (denoted $\mathbb{C}$ ). The integers ( $\mathbb{Z}$ ) are not a field: the only nonzero integers with an integer multiplicative inverse are $\pm 1$ . There are lots of other fields including the $\mathbb{F}_p$ , the fields of integers modulo a prime $p$ . In particular, the field $\mathbb{F}_2 = \{ 0, 1 \}$ is the field with two elements.

Most of the usual things you expect from the algebra of real and complex numbers hold in an arbitrary field. One such thing is the basic algebra of polynomials. Given a field $K$ we can consider $K[X]$ , the collection of all polynomials in the indeterminate $X$ with coefficients in $K$ . The usual stuff still works: a polynomial of degree $n$ has at most $n$ roots in $K$ , roots correspond to linear factors, and we can do polynomial long division.

Now let $f(X) = aX^2 + bX + c$ be a quadratic polynomial with coefficients in $K$ .

If $f(X)$ has a root (say $\alpha$ ) in $K$ then $X - \alpha$ divides $f(X)$ in $K[X]$ . Thus

$f(X) = (X-\alpha)g(X)$

where $g(X)$ must have degree 1. So $g(X) = \beta X + \gamma \in K[X]$ and $g(X)$ clearly has the root $-\gamma/\beta$ . Therefore if a quadratic polynomial has one root in $K$ then it has both of its roots in $K$ .

On the other hand if $f(X)$ has no roots in $K$ we can build a larger field $L$ which contains $K$ and one (and hence both) roots of $f(X)$ . A sketch: Let $\alpha$ be a symbol and set

$L = \{ x + y\alpha : x, y \in K \}$

Define addition and multiplication in the obvious way and by setting
$\displaystyle{\alpha^2 = -\frac{b}{a}\alpha - \frac{c}{a} }$ .
To be explict we have

$(x_1 + y_1\alpha) + (x_2+y_2\alpha) = (x_1 + x_2) + (y_1 + y_2)\alpha$

and

$(x_1 + y_1\alpha) \cdot (x_2+y_2\alpha) = (x_1 x_2 + y_1 y_2 \alpha^2) + (x_1 y_2 + y_1 x_2)\alpha = x_3 + y_3 \alpha$

where $x_3, y_3 \in K$ are calculated by collecting terms and using the relationship above for $\alpha^2$ . Now tedious but elementary arguments show that $L$ is a field containing $K$ and $\alpha \in L$ is a root of $f(X)$ .

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2009 October 5

Dr. Nichtgegeben permalink

As illustration of the construction outlined above, consider the complex numbers. We start with the real numbers and the polynomial $X^2 + 1$ . It has no root in the reals and so we extend things to get a bigger field where the elements have the form $x + y i$ where $x, y$ are real and $i^2 = -1$ .

As another example, consider the field of two elements $\mathbb{F}_2$ and the polynomial $X^2 + X + 1$ . Again this poly has no root in our field so we can build a larger field $L$ where the elements have the form $x + y \alpha$ where $x, y \in \mathbb{F}_2$ and $\alpha^2 = \alpha + 1$ . This then is a field with 4 elements.

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