Fields - Algebraic Elements

Algebraic Elements

Which number is “further” from the integers, $\sqrt{2}+\sqrt{3}$ or $\sqrt{2}$? It intuitively feels like it would be harder to get to $\sqrt2+\sqrt3$ using the integers and the operations we know, like squarerooting, adding, and subtracting. But is there any quantifiable way to explain those intuitions?

This post is to help show what makes a number algebraic. To do this, I need to explain what fields are, and then we can expand the idea to show that some numbers, like $\sqrt{2}+\sqrt{3}$ are “harder” to get than $\sqrt{2}$ from the integers, $\mathbb{Z}$.

What are Fields?

Fields are a unique type of algebraic structure. They are a lot like a group, but instead of just having one operation, they have two. “Addition”, and “Multiplication”. I only put those in quotes because technically they don’t have to behave like the good old addition and multiplication we know, but for the most part, we can assume that they are.

Along with this, fields have distributivity. It says that, \(a(b+c) = ab + ac,\) which means that addition and multiplication have a means of interacting with each other.

We define addition and division the way we expect them to be, and we have them act on our regular numbers, whether they be $\mathbb{Z}$ or $\mathbb{R}$.

Algebraic Numbers and Degrees

Now when we say a number is algebraic in a certain field, we mean that some polynomials with coefficients in that field has a root that is the number we are looking for.

Example

$\sqrt{2}$ is algebraic in $\mathbb{Z}$. Why? Because if $x=\sqrt2$, then $x^2=2$, so a polynomial that we can form with $\sqrt2$ as a root is $f(x)=x^2-2$.

In fact, in the last example, it’s easy to see that $x^2-2$ doesn’t split into any smaller polynomials. This means that $f(x)$ is an irreducible polynomial. In reality, all algebraic numbers of a field have a unique irreducible polynomial that has it as a root.

This fact allows us to get a scale for how algebraic a number can be. We see that the irreducible polynomial that has a root of $\sqrt2$ has a degree (largest power of $x$) of $2$. This motivates us to define the degree of $\sqrt2$ over $\mathbb{Z}$ as $2$.

Now think about the beginning question, what polynomial has a root of $\sqrt2+\sqrt3$? Well a good starting point, like the last one, was to let $x=\sqrt2+\sqrt3$, and try to create a polynomial with integer roots out of it. The process is as follows:

\[\begin{align*} x &= \sqrt2+\sqrt3 \\ x- \sqrt2 &= \sqrt3 \\ x^2-2\sqrt2x + 2 &= 3 \\ x^2 -1 &= 2\sqrt2x \\ x^4-10x^2+1&=0 \end{align*}\]

Our desired polynomial is $x^4-10x^2+1$. Checking that this is irreducible is important, but I won’t cover it here, (see Einsenstein’s Irreducibility Criterion). Once we know it is irreducible, then we can conclude that $\sqrt2+\sqrt3$ has degree $4$ in $\mathbb{Z}$.

Well then, since we need a higher degree to find a polynomial with a root of $\sqrt2+\sqrt3$, there is a mathematical justification for why it is “harder” to get $\sqrt2+\sqrt3$ over $\sqrt2$.

Bonus: Transcendentals

Transcendental numbers are often discussed when talking about $\pi$ or $e$. The definition of transcendental number is simple: a number is transcendental in a field $F$ if it isn’t algebraic. What that’s saying is that there is no polynomial that exists that has $\pi$ or $e$ as a root. Proofs for these are quite difficult, but a google search may be worth the time to look at them.