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Is a field just a commutative ring?

My algebra professor didn't give a very wide introduction to this algebraic structure, and I did not get a real grasp of what a field is.

We're studying polynomials of $\mathbb{K}[x]$ and he keeps repeating $\mathbb{K}[x]$($\Bbb{C}[x], \Bbb{R}[x] $ or $\Bbb{Q}[x]$) are fields so they have $y$ or $z$ property, but I don't get the difference between them and a commutative ring.

YoTengoUnLCD
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  • A field F is a commutative ring in which every nonzero element of F is a unit. – Mr.Fry May 08 '15 at 02:33
  • No, the polynomials do not form a field. You might have mistaken the notation for $\mathbb{K}\left[x\right]$ where $x$ is a given algebraic element of a field extension of $\mathbb{K}$ (rather than a polynomial indeterminate); in this case, $\mathbb{K}\left[x\right]$ is a field (and is not a polynomial ring). – darij grinberg May 08 '15 at 02:33
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    It is very likely that the rational functions are meant, in which case you should have typed round parentheses instead of brackets – Will Jagy May 08 '15 at 02:36

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All fields are (nonzero) commutative rings, but not all commutative rings are fields.

The special property that distinguishes fields from commutative rings is that they contain a nonzero multiplicative inverse for every nonzero element.

Sam
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Newb
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  • I would say that a commutative ring is non-zero also. Of course, in the definitions, one is not prohibited by having $0=1$, but in commutative algebra rings are not trivial. Modules can be trivial but not rings themselves, I cannot recall a time when we actually deal with a zero ring. – Nicolas Bourbaki May 08 '15 at 02:35
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    @NicolasBourbaki Sure, but the zero ring is trivially commutative, right? – Newb May 08 '15 at 02:36
  • Could you give an example of a commutative ring that is not a field? I'm having trouble visualizing it. – YoTengoUnLCD May 08 '15 at 02:37
  • @Newb Of course, I did not say you were wrong about anything. I was only making a comment about convention. Are you rings trivial or not? My rings are never trivial and I cannot think of a time when we allowing zero rings are useful in commutative algebra. – Nicolas Bourbaki May 08 '15 at 02:37
  • @YoTengoUnLCD, the ordinary integers taken modulo $15$ are a commutative ring with multiplicative identity, but there are zero-divisors – Will Jagy May 08 '15 at 02:38
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    @YoTengoUnLCD The integers, with the operations $(\cdot, +)$ form a commutative ring. But in this structure, there are no multiplicative inverses yet, so it's impossible to solve a problem like $2x = 1$. We need multiplicative inverses ("division") to solve an equation like that. When we add those multiplicative inverses to our set of the integers, we have a field (the rational numbers). – Newb May 08 '15 at 02:40