Field Theory and Galois

Recall that the complex numbers are numbers of the form $a+bi$ where $a, b \in \mathbb{R}$, i.e. $a$ and $b$ are real numbers, and the symbol $i$ satisfies the equation $i^2 + 1 = 0$. The complex numbers form a field, i.e. we can do all the usual operations we expect to be able to do with numbers, like addition, subtraction, multiplication and division.

Another property of the complex numbers that allows us to call them a field is that multiplication is commutative, i.e. $ab = ba$ for all complex numbers $a$ and $b$.

Finally, we note that the complex numbers satisfy the distributive rule, i.e. $a(b+c) = ab+ac$ for all complex numbers $a$, $b$ and $c$.

In fancy mathematical language, we can say that the complex numbers are isomorphic to the field $\mathbb{R}[X]/(X^2+1)$, the quotient of the polynomial ring $\mathbb{R}[X]$ by the maximal ideal $(X^2+1)$. But you don't have to understand what this means at this point.

Notice that every complex number $z = a+bi$ is described completely by specifying two real numbers, $a$ and $b$. In fancy mathematical language, way say that the complex numbers form an extension of the field of real numbers of degree $2$.

This allows us to plot complex numbers on the two dimensional cartesian plane, the complex number $z = a+bi$ corresponds to the point on the cartesian plane $(a, b)$.

We call such a graph an Argand diagram, named after the mathematician Argand, although he wasn't the first to describe it!

It now becomes clear that we can associate a length to every complex number - this is the length from the origin of the Argand diagram to the point corresponding to that complex number.

This length is a non-negative real number, and we call the length of a complex number the modulus of that number.

We denote the modulus of the complex number $z = a+bi$ by $|z| = \sqrt{a^2 + b^2}$.

We can also associate with every complex number a direction, this is the direction from the origin of the Argand diagram. We call this direction the argument of the complex number, and denote the argument of the complex number $z$ by $\arg(z)$. We usually measure the argument of a complex number in radians, and measure counterclockwise, starting from the $x$ axis.

We have to be careful though! If we rotate the complex number a full turn around the origin, we arrive back at where we started, thus we come back to the same number.

By convention, the argument of a complex number is usually chosen to lie in the interval $(-\pi, \pi]$.

We now introduce a result that pertains to the modulus.

For any complex numbers $z$ and $w$, we have $|zw| = |z||w|$. This means that we can either multiply $z$ and $w$ as complex numbers and then take the modulus, or we can take the modulus of $z$ and $w$ separately and then multiply them as real numbers, to get the same result.

We can also think about this in terms of a geometrical effect of multiplying the complex number $z$ by the complex number $w$ - the modulus of $z$ increases (or decreases, if $|w| < 1$) by a factor of $|w|$!

Can you prove this result?

We now introduce a result pertaining to the arguments of complex numbers.

For any complex numbers $z$ and $w$, we have $\arg(zw) = \arg(z) + \arg(w)$. Notice that we may have to add or subtract $\pi$ to get the result within the conventional range of $(-\pi, \pi]$.

Geometrically, this says that an effect of multiplying the complex number $z$ by the complex number $w$ is to add the argument of $w$ onto the original argument of $z$.

We now concentrate on complex numbers with modulus equal to $1$. These lie on the unit circle on the Argand diagram.

We use the symbol $\mathrm{e}^{i\theta}$ to denote the complex number with argument $\theta$ and unit modulus. Thus $|\mathrm{e}^{i\theta}| = 1$ and $\arg(\mathrm{e}^{i\theta}) = \theta$ (up to some multiple of $2\pi$).

By what we have learnt so far we know that the geometrical effect of multiplication by $\mathrm{e}^{i\theta}$ is to rotate by $\theta$ radians, while leaving the modulus the same.

In particular, $\mathrm{e}^{i\theta}\mathrm{e}^{i\phi} = \mathrm{e}^{i(\theta+\phi)}$. Think about this carefully! Perhaps draw a sketch, labelling all three angles.

Suppose $|z| = r$ and $\arg(z) = \theta$, then $z = r\mathrm{e}^{i\theta} = r(\cos(\theta)+i\sin(\theta))$.

Say we also have $|w| = \rho$ and $\arg(w) = \phi$, so that $w = \rho\mathrm{e}^{i\phi}$. Then $|zw| = |z||w| = r\rho$, and $\arg(zw) = \theta + \phi$. Thus $zw = r\rho\mathrm{e}^{i\theta}\mathrm{e}^{i\phi} = r\rho\mathrm{e}^{i(\theta+\phi)}$.

As $\mathbb{C}$ is a field extension of $\mathbb{R}$of degree $2$, $\mathbb{C}$ forms a vector space over $\mathbb{R}$ of dimension $2$. A standard basis for this vector space would be $\{1, i\}$.

Multiplication by an element $\alpha \in \mathbb{C}$ gives us a linear map $\mathbb{C} \to \mathbb{C}$. We denote the determinant of this map as $N(\alpha)$, the norm of $\alpha$.

By what we have said so far, in the standard basis mentioned above, the matrix representing non-zero $\alpha$ is a rotation matrix multiplied by $|\alpha|$.

As $|\mathbb{C}:\mathbb{R}| = 2$, there can be at most $2$ $\mathbb{R}$-homomorphisms in $\mathrm{Hom}_{\mathbb{R}}(\mathbb{C})$. We have one nontrivial homomorphism - complex conjugation, that sends $a+bi \to a-bi$.