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The following comes from Baby Rudin:

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Is there any special reason as to why Rudin makes $1\neq 0$ a requirement? I've seen he uses this fact in a few elementary proofs below, the only consequence of this is excluding the $\{0\}$ field, right?

YoTengoUnLCD
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    It's not just Rudin, but a common requirement. And yes, its purpose is to exclude the trivial one-element field. – vadim123 Dec 12 '15 at 00:43
  • The subfield generated by $1$ is called the prime field of a given field. If this is finite, it is isomorphic to $\Bbb Z/p\Bbb Z$, for a prime $p$. If we allowed $0 = 1$, the resulting "prime field" would be isomorphic to $\Bbb Z/\Bbb Z$, and one hesitates to call $1$ a prime number. – David Wheeler Dec 12 '15 at 00:46
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    @HenningMakholm: I think you are right about the duplication. Interestingly, the closely related question "why are maximal ideals required to be proper ideals" doesn't seem to get asked on MSE? And no-one seems to worry about the fact that the trivial ring has no maximal ideals. Perhaps once people have got to ideals, they are so happy with the definition of field, that all is then clear. – Rob Arthan Dec 12 '15 at 01:51
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    Main reason if you ask me: You want the theorem "$R/I$ is a field $\Longleftrightarrow$ $I$ is a maximal ideal" to hold (where $R$ is a commutative ring and $I$ is an ideal). – darij grinberg Dec 12 '15 at 02:04

2 Answers2

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In algebraic geometry, fields loosely correspond to "points." The zero ring does not correspond to a point: instead, it corresponds to the empty set.

More practically, it just turns out that the zero ring doesn't behave like a field, so you shouldn't call it a field. For example, there's a unique module over the zero ring, namely the zero module. Modules over a field should be vector spaces, and there should be one for every possible dimension.

One way to write down the definition of field so that it naturally excludes the zero ring is that a field is a commutative ring with exactly two ideals. The zero ring only has one ideal. See also too simple to be simple.

Qiaochu Yuan
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Even simpler than Qiaochu Yuan's nice definition involving ideals, it's convenient and memorable to define a field as a ring whose non-zero elements form a group under multiplication. As groups can't be empty, that excludes the possibility that $0 = 1$.

Rob Arthan
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  • I realise that this is a very old post, but shouldn't the definition be "a field is a commutative ring whose non-zero elements form a group under multiplication", or "a field is a ring whose non-zero elements form an abelian group under multiplication". Although perhaps the definition of "ring" that you are using already stipulates that all rings are commutative. – Joe Feb 18 '22 at 16:58
  • @Joe: in French usage, fields are not required to be commutative (i.e., "corps" means what English speakers call "division ring", not "field"). So there isn't any universal agreement on whether fields are commutative. I leave it to you to insert the word "commutative" wherever you fell it is appropriate. The issue under discussion has nothing to do with commutativity. – Rob Arthan Feb 19 '22 at 22:10