Thou who art on math.stackexchange.com, please throw some light on this definition of fields I read somewhere..

A field is a commutative ring with $0 \ne 1$, and every non-zero element is invertible.

I know about fields. They are rings with inveritble elements, except the additive identity which doesn't have an inverse with respect to * operation.

I am not getting what do they mean by $0 \ne 1$ ?

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    $\begingroup$ they mean a Field has to have at least two elements, and all non-zero elements have to be invertible, For instance {0} is not a Field, but {0,1} is $\endgroup$ – Deven Ware May 6 '12 at 1:37
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    $\begingroup$ The condition $0 \neq 1$ is typically written to avoid trivial fields i.e. field with just one element. $\endgroup$ – user17762 May 6 '12 at 1:38
  • $\begingroup$ Oh Math-magicians.. thanks!! I understand it now. $\endgroup$ – vidit May 6 '12 at 1:42
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    $\begingroup$ (Someone should post an answer so that this question is lot left barren) $\endgroup$ – Mariano Suárez-Álvarez May 6 '12 at 1:45

Here's the explicit description of what a field is:

A field is an ordered triple $(F,+,\times)$, where $F$ is a set, $+$ and $\times$ are binary operations (functions $F\times F\to F$; we use infix notation and write $a+b$ instead of $+(a,b)$, and similarly with $\times), such that:

  1. $+$ is associative: $(a+b)+c = a+(b+c)$ for all $a,b,c\in F$.
  2. There is an element $0\in F$ such that for all $a\in F$, $a+0=0+a=a$.
  3. For every $a\in F$ there exists $b\in F$ such that $a+b=b+a=0$.
  4. $+$ is commutative: for all $a,b\in F$, $a+b=b+a$.
  5. $\times$ is associative: for all $a,b,c\in F$, $a\times(b\times c)=(a\times b)\times c$.
  6. $\times$ distributes over $+$ on the left: for all $a,b,c\in F$, $a\times(b+c) = (a\times b)+(a\times c)$.
  7. $\times$ distributes over $+$ on the right: for all $a,b,c\in F$, $(b+c)\times a = (b\times a) + (c\times a)$.
  8. There exists an element $1\in F$ such that for all $a\in F$, $1\times a = a\times 1 = a$.
  9. For every $a\in F$, if $a\neq 0$ then there exists $b\in F$ such that $a\times b = b\times a = 1$.
  10. $\times$ is commutative: for all $a,b\in F$, $a\times b = b\times a$.
  11. $0\neq 1$.

Suppose we only consider the first 10 conditions. Then we can take $F=\{0\}$, and define $+$ to be $0+0=0$, and $\times$ to be $0\times 0 = 0$. Then $(\{0\},+,\times)$ staisfies the first 10 axioms, if we interpret $1$ to mean $0$. That is, (8) above does not require that the element we call $1$ be different from the element we called $0$ in (2). Without the requirement that they be different (which is given in (11)), $(\{0\},+,\times)$ is a perfectly fine "field".

Note that if $0=1$ in a field, then automatically we get that every element of the field is $0$; that is, we just have one element: by (2), we have $0+0=0$. By (6), for all $a\in F$ we have $a\times 0 = a\times (0+0) = (a\times 0)+(a\times 0)$. Then using (3) there exists $b$ such that $b+(a\times 0)$; adding to both sides of the equation we get $$0 = b+(a\times 0) = b+\Bigl((a\times 0)+(a\times 0)\Bigr) = \Bigl(b+(a\times 0)\Bigr)+(a\times 0) = 0+(a\times 0) = a\times 0,$$ so $a\times 0 = 0$ for all $a\in F$. But since $0=1$, by (8) we also have $a=a\times 1 = a\times 0 = 0$. So every element of $F$ is $0$.

So we just have two possibilities: either everything is $0$, or else $0\neq 1$.

Why do we object to the first possibility? Turns out that this object behaves very differently from every other field; a lot of results become too cumbersome to state when we keep having to exclude the field with $0=1$ explicitly. And for some of the most important uses of fields, the field with $0=1$ is useless. So rather than keep dragging it along, not using it, and excluding it from most results explicitly, we just kick it out of the club by requiring that $0\neq 1$ hold in a field.

Nonetheless, it turns out that there are some situations where the "one element field" is useful! This is called the field with one element and plays an important role in noncommutative geometry and as a unifying concept.

  • $\begingroup$ The additive identitiy ... is it always $0$? $\endgroup$ – vidit May 6 '12 at 14:43
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    $\begingroup$ @vidit: It is traditional in rings to call the additive identity "0" unless we have good reasons to do otherwise; it is also traditional to call the multiplicative identity "1". But these are the names we give these objects. We can give the real numbers a ring structure completely different from the usual one; for example, define $a\oplus b = a+b+1$, and $a\otimes b = ab+a+b$; you can verify that this makes $(R,\oplus,\otimes)$ into a ring, but in this case the additive identity happens to be the real number $-1$ (and the multiplicative identity is the real number $0$). $\endgroup$ – Arturo Magidin May 6 '12 at 19:10

First look at the axioms for a commutative ring $R$. It must be associative with respect to addition and multiplication, there must be multiplicative and additive identities, as well as distributivity, additive inverses, and we need commutativity with addition and multiplication.

For a field we need all of that as well multiplicative inverses (for non-zero elements) which explains the second part of the statement. When we also require that $0\neq 1$, this is simply to prevent entirely trivial fields that would normally have to be dealt with as separate cases. Thus $\{0\}$ is not a field, but $\{0,1\}=\mathbb{F}_2$ is a field.


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