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I was introduced to the definition of a field today, as a communtative ring with two operations, like $\Bbb{R} = \langle R, +, -, \cdot, ^{-1} \rangle$ and all the usual axioms; commutativity, associativity, identity & inverse elements, but there was a remark in the axioms that said the identity element of addition must be distinct from then identity element of multiplication, $ 0 \neq 1$. Is there a reason why this must be, and is there an example of a ring where this is not the case?

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    $\begingroup$ If $0 = 1$ in a ring $R$, then $0 = 0x = 1x = x$ for all $x\in R$; that is, $R = 0$. $\endgroup$ – anomaly Jun 29 '16 at 16:25
  • $\begingroup$ Suppose that $1=0$. Then $x=1\cdot x = 0\cdot x = (0+0)\cdot x = 0x + 0x = 1x+1x=x+x$ implying that $x=0$ $\endgroup$ – JMoravitz Jun 29 '16 at 16:25
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The zero ring, a ring containing only one element, necessarily has $0 = 1$, since this ring satisfies all the ring axioms (except $0 \neq 1$), but only has one element. The zero ring is a bit pathological, so rather than writing "Let $R$ be any ring other than the zero ring", we exclude the zero ring via the "$0 \neq 1$" condition.

There is also a notion "field with one element". Note that even if one works in a setting where the field with one element makes sense, this object is not a field, for the same axiomatic reason as is used to exclude the zero ring.

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The identity element under addition is denoted by $0$. It has the property that $0*a = 0 = a*0$ for all $a\in R$.

The identity element under multiplication is denoted $1$. It has the property that $1*a = a = a*1$ for all $a \in R$.

If $0 = 1$, then for all $a \in R$,

$$ 0 = 0*a = 1*a = a.$$

So $R = \{0\}$. We don't like to call this a field as it doesn't behave like one.

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If $0 = 1$, then for any $x$ in the ring: $$x = x.1 = x.0 = 0$$ and therefore there is only one element in the ring.

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