If something is algebraic in a field, what does it mean?

I don't know the correct phrasing. An element $a \in K$ is algebraic over $F$. Please can someone give some correct phrasing with a simple example.


$a$ is algebraic over $F$ just when it is a root of a polynomial with coefficients from $F$. So $\sqrt{2}$ is algebraic over $\mathbb{Q}$ since it's a root of the polynomial $x^2 - 2$.

Edit: More examples, more context.

First a small finite example - essentially, the smallest example in the family suggested by @fleablood 's comment.

Here are the addition and multiplication tables for the four element field:

+ | 0 1 a b       * | 0 1 a b
-------------     -------------
0 | 0 1 a b       0 | 0 0 0 0
1 | 1 0 b a       1 | 0 1 a b
a | a b 0 1       a | 0 a b 1
b | b a 1 0       b | 0 b 1 a

Then $a$ is a root of the polynomial $x^2 + x + 1$ with coefficients in the two element subfield $F = \{0,1\}$ so it's algebraic over that subfield.

In order to understand what "algebraic" means you should also consider situations where it fails. (That's true for any new mathematical concept.) To see examples you have to move beyond finite. If a field $K$ containing a field $F$ is finite dimensional (not necessarily finite) as a vector space over $F$ then every element $a$ of $K$ is algebraic over $F$ since the set $\{1, a, a^2, \ldots \}$ must be linearly dependent.

As @ZacharySelk notes in his answer, $\pi$ is not algebraic over the rationals. But that's pretty deep. It's easier to prove that Liouville numbers (https://en.wikipedia.org/wiki/Liouville_number) are not algebraic.

There are examples that don't depend on analysis. For any field $F$ (think about the rationals) let $K$ be the field whose elements are quotients of polynomials with coefficients from $F$. Then the element $x$ of $K$ is not the root of any polynomial with coefficients in $F$, so it's not algebraic.

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    $\begingroup$ Isnt $x^2-2$ the minimum polynomial of $\mathbb Q ( \sqrt2)$ over $\mathbb Q$? $\endgroup$ – snowman May 15 '16 at 0:52
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    $\begingroup$ It is, but it has nothing to do with it. $\endgroup$ – B. Pasternak May 15 '16 at 0:53
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    $\begingroup$ I am confused mainly because you didn't define everything from scratch. I am guessing: $Q \subseteq F$ so $F$ is a field extension over $Q$. We have that $x^2-2 \in Q[x]$ and we have that $\sqrt2 \in F$ is a root of this, meaning that it is algebraic. Is this correct? $Q$ is the rationals. $\endgroup$ – snowman May 15 '16 at 1:29
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    $\begingroup$ Not nesc. F can be any field at all. But the term a must be in some extension so that the operations with a are defined. (Field elements are trivially algebraic). $\endgroup$ – fleablood May 15 '16 at 2:07
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    $\begingroup$ @snowman I think Ethan did define almost everything from scratch: we say $a$ is algebraic over $F$ if $a$ is a root of some polynomial with coefficients from $F$. What's not from scratch, here? $\endgroup$ – Noah Schweber May 15 '16 at 2:15

It means it's the root of a polynomial with coefficients in $F$.

For example $\sqrt{2}$ is algebraic over $\Bbb{Q}$ where $\pi$ isn't.

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    $\begingroup$ Or maybe instructive to point out that $\pi$ is algebraic over $\Bbb R$ but not over $\Bbb Q$. $\endgroup$ – MJD May 15 '16 at 2:36
  • $\begingroup$ @MJD, didn't you post a nice explanation of "algebraic" quite a while back? :) $\endgroup$ – J. M. is a poor mathematician May 15 '16 at 2:46

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