# Why do people study algebraic extension?

Yesterday, I learned Kronecker’s theorem and a finite extension. And now I’m studying the next chapter, Algebraic extension.

I think the next theorem shows how important algebraic extension is

Let $E$ be an extension field of a field $F$. Let $α$ be an element of $E$. If $α$ is algebraic over $F$, there exists a unique monic irreducible polynomial $p(x)$ in $F[x]$ such that $p(α)=0$ in $E$.

But I’m wondering there is another important reason why we should study algebraic extensions.

• Ultimately the theory leads to such things as the impossibility of trisecting arbitrary angles with compass and straightedge, and the impossibility of solving arbitrary higher degree polynomials with radicals. Is that important enough? – John Coleman Apr 3 '18 at 0:55
• Oh.. Thanks! My book just gave me the definition of algebraic extension so I was curious about it. I should keep on reading my textbook! I want to learn those things ..! Thanks ! – ylh0501 Apr 3 '18 at 1:02
• – Asaf Karagila Apr 3 '18 at 15:22
• There are many algebraically closed fields which are extension fields of a given field. To obtain some uniqueness, we need to consider algebraic extension. – Hunter Liu Oct 9 '18 at 23:03

I think the original motivation for studying field extensions was, as in the theorem you stated, to solve polynomials. One of the big results after a few lectures of algebraic field extensions is that every field can be embedded into a unique algebraically closed field, called the algebraic closure.

Actually, solving equations is really the motivation for all the historical expansions of the concept "number." Think of it this way: we have the natural numbers, $1,2,3,\ldots$ and we can solve equations like $x+2 = 4$. But then we can pose equations like $x+ 2 = 2$ and $x+4 = 2$, so we want to extend our number system to include solutions to these, so we add zero and negative integers.

Then we notice we can pose equations like $2x = 4$, which has a nice solution $x=2$, and also $4x = 2$, which doesn't have a solution in the integers. So again, we extend our number system again to include things like $\frac 12$. Now we can solve any linear equation $ax+b = 0$ where $a,b \in \mathbb{N}$.

We also have equations like $x^2 - 2 = 0$, and so we start adding irrational numbers like $\sqrt{2}$. We can complete the rational numbers to form the field of real numbers $\mathbb{R}$. But still, we have equations we can't solve, like $x^2 + 1 = 0$. To get a solution of this, we add the number $i = \sqrt{-1}$ and get the complex numbers. Going from $\mathbb{R}$ to $\mathbb{C}$ this way is an algebraic field extension.

As the theorem you stated says, this procedure is much more general than just $\mathbb{R}$ to $\mathbb{C}$: it says that given any abstract field $F$ and any polynomial equation with coefficients from that field, we can enlarge the "number system" that is $F$ to include a solution.

EDIT: As was pointed out in a comment, the algebraic closure of a field is only unique up to a non-canonical isomorphism.

• This might be overdue, but I just wanted to emphasize that in general there is no canonical choice of an algebraic closure. It is unique up to non-canonical isomorphisms, and its existence in general requires the axiom of choice. I like the answer very much, in any case ;) – 57Jimmy Apr 3 '18 at 10:08
• To nitpick, $\mathbb{C}$ is a very different beast than what you get by algebraic extension, $\overline{\mathbb{Q}}$. Almost all numbers in $\mathbb{C}$ are not in $\overline{\mathbb{Q}}$, such as $\pi$ and $e$. You never get to $\mathbb{R}$ or $\mathbb{C}$ from algebraic extensions of the naturals. – orlp Apr 3 '18 at 15:48
• @orlp Very true. Going from $\mathbb{Q}$ to $\mathbb{R}$ is not so much about about solving equations as having limits of Cauchy sequences. I didn't mention this because it seemed a bit of tangent. – Joshua Ruiter Apr 3 '18 at 15:56
• But both $\overline {\Bbb Q}$ and $\Bbb C$ are algebraicly closed extensions of $\Bbb Q$, so it is not true that any field has a unique algebraically closed extension. – Paul Sinclair Apr 3 '18 at 16:31
• As orlp noted, $\mathbb{C}$ is not an algebraic extension of $\mathbb{Q}$. It is a transcendental extension. The uniqueness property says that there is a unique (up to isomorphism) algebraic extension of a field $F$ that is algebraically closed. – Joshua Ruiter Apr 3 '18 at 22:38

It's important in algebraic number theory. Without going into too much detail, many results about the ring $\mathbb{Z}$ of integers are easier to obtain and understand when we generalize the notion of an integer. This is done by replacing the field $\mathbb{Q}$ of rational numbers with a field $K$ of the form $\mathbb{Q}(a)$, where $a$ is a complex number which is algebraic over $\mathbb{Q}$.

For the field $\mathbb Q$ of rational numbers, the ring of integers is of course $\mathbb{Z}$.

For the field $\mathbb{Q}(i) = \{ a + bi : a, b \in \mathbb{Q}\}$, the ring of integers is $\mathbb{Z}[i] = \{a + bi : a, b \in \mathbb{Z} \}$. This is also called the ring of Gaussian integers.

More generally, every field $K$ containing $\mathbb{Q}$ of the form $\mathbb{Q}(a)$ as above, has its own "ring of integers."

Here is a classical theorem about prime numbers:

Theorem: Let $p \geq 3$ be a prime number. There exist integers $x, y \in \mathbb{Z}$ such that $p = x^2 + y^2$ if and only if $p - 1$ is divisible by $4$.

This theorem was known before people were working with the ring $\mathbb{Z}[i]$, but it is much easier to prove using the ring $\mathbb{Z}[i]$. Basically, one defines the notion of an element of $\mathbb{Z}[i]$ being prime. Then the question becomes to show which prime numbers in $\mathbb{Z}$ remain prime when regarded as an element of $\mathbb{Z}[i]$.

You can teach a computer how to calculate with extensions field "exactly". If you see the $\sqrt{2}$ as an infinite decimal, your computer can only compute approximations (in finite time). But if you teach the computer that it is a pair $(0,1)$, with some algebraic properties, it will be able to give back exact answers to your calculations that involve $\sqrt{2}$.