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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.

snowman
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2 Answers2

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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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$a$ 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.

Ethan Bolker
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    Isnt $x^2-2$ the minimum polynomial of $\mathbb Q ( \sqrt2)$ over $\mathbb Q$? – snowman May 15 '16 at 00:52
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    It is, but it has nothing to do with it. – B. Pasternak May 15 '16 at 00:53
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    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. – snowman May 15 '16 at 01:29
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    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). – fleablood May 15 '16 at 02:07
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    @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? – Noah Schweber May 15 '16 at 02:15
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    @snowman : To understand the examples, I think you need to say not that $\mathbb Q\subseteq F$, but rather than $\mathbb Q= F$, and $\mathbb R = K$. The fact that $x^2 - 2\in\mathbb Q[x]$ means, not just that they are "algebraic" but that they are _algebraic over $\mathbb Q$. $\qquad$ – Michael Hardy May 15 '16 at 02:46
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    F doesn't have to have anything to do with Q. Any field will do. Then if you have a polynomial with F coeficients, a is algebraic if plugging a into the polynomial yields the 0 element. No every element in the field is trivially algebraic, so we have to somehow deal with what a is if a isn't in the field and how we use field operations on a. This is simple if a is in a field extension. – fleablood May 15 '16 at 07:27
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    Let F = Z_7 a field under multiplication. $x^2 -3$ doesn't have a solution. If we add an element poo such that $poopoo =3$ we can,with effort see we've created a field F with 14 elements and $Z_7 \subset F$. (Note: $1/poo = 5poo$.) Then $\poo\not \in Z_7$ and poo is algebraic over $Z_7$. That's an example of an algebraic element that has nothing to do with Q. – fleablood May 15 '16 at 07:42
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    @MichaelHardy OK just to confirm, $\sqrt2$ is algebraic over $\mathbb Q$ because we can find some polynomial $f \in \mathbb Q [x]$ such that $\sqrt2$ is a root of $f$. We can easily choose $f$ to be $x^2-2$ like you guys are saying? Is this correct? – snowman May 15 '16 at 12:50
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    @snowman Yup, that's right – Simply Beautiful Art May 15 '16 at 13:52
  • What does algebraic data types mean? In other words, what’s algebraic about them? – Behrang Saeedzadeh Oct 16 '18 at 14:16
  • @Behrang We use the adjecttive "algebraic" for algebraic numbers (over a field) because they can be defined and studied using a finite number of algebraic operations - addition, subtraction, multiplication and division and dealing with roots of polynomials (even when if there is no explicit formula for the roots), When $F$ is the field of rational numbers the real numbers that aren't algebraic, like $\pi$, are called "transcendental". You need limiting arguments to characterize and study them. – Ethan Bolker Oct 16 '18 at 16:53
  • @EthanBolker can I email you to discuss algebraic data types? In particular, I want to know what's algebraic about them? Or similarly, what is algebraic about relational algebra? I think we can't explain these algebras using roots, coefficients, and polynomials but I might be wrong. – Behrang Saeedzadeh Oct 16 '18 at 23:30
  • @Behrang Those are entirely different uses of the very general adjective "algebraic" and have little to do with the way mathematicians use the word. The designation carries no formal or precise meaning. I'm afraid I can't help more than the links you offer. – Ethan Bolker Oct 17 '18 at 01:26
  • @EthanBolker This has been keeping my mind busy for a few weeks now. There should be a reason why they have called it "algebraic data types" instead of "something else data types", say, "logical data types", "combinatoric data types", "differential data types", etc. There should be a way to look at a programming language such as Java and say "Java does not support algebraic data types" or saying that "OCaml supports algebraic data types". It is just not clear enough how we can make such distinguishments. – Behrang Saeedzadeh Oct 18 '18 at 00:21