Fields

The following objects are really important in mathematics:

So what do they have in common? Well, a lot, but the important part is that they are all fields. So here we go: A field is a set along with operations + and \cdot such that the elements in the field satisfy:
  • Closure: If a and b are in the field, so are a+b and a\cdot b
  • + and \cdot  are associative. That is, (a+b)+c=a+(b+c) and (a\cdot b)\cdot c)=a\cdot(b\cdot c)
  • a+b=b+a, and a\cdot b=b\cdot a
  • a\cdot(b+c)=a\cdot b+a\cdot c
  • There is an “additive identity” element 0 such that a+0=a
  • For each a there is a unique “additive inverse” (-a) such that a+(-a)=0
  • There is a “multiplicative identity” 1 such that a\cdot 1=a
  • For each a (other than 0) there is a unique “multiplicative inverse” a^{-1} such that a\cdot a^{-1}=1. (i.e., don’t divide by zero!)
  • EDIT: Also, 0\neq 1
Hmmm, this is a lot of rules. How are you going to remember them? Well, you’ve been practicing them your entire life. Most of the arithmetic you have done has been in one of the fields \mathbb Q, \mathbb R,or \mathbb C. You can verify for yourself that these are fields. This is just an abstract formalization of what you already know.
A few more points to think about:
  • \mathbb Z, the set of integers is not a field. There is no integer n for which 2\cdot n=1, so 2 has no multiplicative inverse.
  • We can define subtraction and division to be what they should be a - b means a + (-b), where (-b) is the additive inverse of b. \frac ab=a\cdot b^{-1} where b^{-1} is the multiplicative inverse of b.
  • Finite sums/products only. Sorry, I don’t make the rules.
  • Can you prove that 0\cdot a=0 for every a in an arbitrary field?
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32 Responses to Fields

  1. Benji says:

    I think of the axioms this way. First, addition and multiplication are functions: each takes two elements of the field and outputs a field element. (I do not think of closure as an axiom, but that is just point of view.) Next, four axioms for each of the two operations: associativity, commutativity, identity, and inverse. Finally, the two axioms that connect the two operations: distributivity and the forgotten axiom.

    Guess what? You forgot one. You are not alone.

    To rule out the case of a field with one element, I add the forgotten axiom: 0 is not equal to 1. Yes, there are some mathematicians who do not think this is part of the definition of a field, but they are a very small minority.

    • soffer801 says:

      Thanks Benji!

      I did forget that, and you are correct. I fixed it now.

    • It was once pointed out to me that the assumption that for fields, 0 and 1 are different, is exactly the same assumption that 1 is not prime.

      This is true in greater generality, but you can see it following from Andy’s next post, where he mentions that Z/pZ is a field for all prime p.

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  10. Luqing Ye says:

    \forall a\in\mathbb{F}, a\cdot 0+a\cdot 1=a\cdot (0+1)=a\cdot 1=a.

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