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Is a commutative ring a field? A set equipped with addition and multiplication which is abelian over those two operations and it holds distributivity of multiplication over addition?

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    $\begingroup$ No, all non-zero element in a field must be invertible, which need not hold in an arbitrary ring. $\endgroup$ Commented Jun 28, 2012 at 9:57
  • $\begingroup$ So this is the property that makes those two sets different $\endgroup$
    – curious
    Commented Jun 28, 2012 at 9:59
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    $\begingroup$ The other difference is that in a field $1$ must be different from $0$. $\endgroup$ Commented Jun 28, 2012 at 11:05
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    $\begingroup$ Why did you think they were the same? $\endgroup$
    – lhf
    Commented Jun 28, 2012 at 11:21
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    $\begingroup$ I made the very relaxed assumption that a field is just a ring but with the commutativity in multiplication as well which is the case of the commutative ring $\endgroup$
    – curious
    Commented Jun 28, 2012 at 11:31

2 Answers 2

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A key difference between an ordinary commutative ring and a field is that in a field, all non-zero elements must be invertible. For example:

$\Bbb{Z}$ is a commutative ring but $2$ is not invertible in there so it can't be a field, whereas $\Bbb{Q}$ is a field and every non-zero element has an inverse.

Examples of commutative rings that are not fields:

  1. The ring of polynomials in one indeterminate over $\Bbb{Q}, \Bbb{R}$, $\Bbb{C}$, $\Bbb{F}_{11}$, $\Bbb{Q}(\sqrt{2},\sqrt{3})$ or $\Bbb{Z}$.

  2. The quotient ring $\Bbb{Z}/6\Bbb{Z}$

  3. $\Bbb{Z}[\zeta_n]$ - elements in here are linear combinations of powers of $\zeta_n$ with coefficients in $\Bbb{Z}$ (In fact this is also a finitely generated $\Bbb{Z}$ - module)

  4. The direct sum of rings $\Bbb{R} \oplus \Bbb{R}$ that also has the additional structure of being a 2-dimensional $\Bbb{R}$ - algebra.

  5. Let $X$ be a compact Hausdorff space with more than one point. Then $C(X)$ is an example of a commutative ring, the ring of all real valued functions on $X$.

  6. The localisation of $\Bbb{Z}$ at the prime ideal $(5)$. The result ring, $\Bbb{Z}_{(5)}$ is the set of all $$\left\{\frac{a}{b} : \text{$b$ is not a multiple of 5} \right\}$$ and is a local ring, i.e. a ring with only one maximal ideal.

  7. I believe when $G$ is a cyclic group, the endomorphism ring $\textrm{End}(G)$ is an example of a commutative ring.

Examples of Fields:

  1. $\Bbb{F}_{2^5}$

  2. $\Bbb{Q}(\zeta_n)$

  3. $\Bbb{R}$

  4. $\Bbb{C}$

  5. The fraction field of an integral domain

  6. More generally given an algebraic extension $E/F$, for any $\alpha \in E$ we have $F(\alpha)$ being a field.

  7. The algebraic closure $\overline{\Bbb{Q}}$ of $\Bbb{Q}$ in $\Bbb{C}$.

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    $\begingroup$ Nitpick: in #5, if $X$ has only one point, then $C(X)$ is a field. $\endgroup$ Commented Jun 28, 2012 at 13:04
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    $\begingroup$ The first line does not compile for me. Is it just me (+ 1:30am)? $\endgroup$
    – Asaf Karagila
    Commented Jun 28, 2012 at 22:27
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    $\begingroup$ Fine on the iOS $\endgroup$ Commented Jun 28, 2012 at 22:37
  • $\begingroup$ @AsafKaragila I'm on chrome with Ubuntu 11.04 and it loads for me. $\endgroup$
    – user38268
    Commented Jun 28, 2012 at 22:57
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    $\begingroup$ How about we use "parse" for engrish [sic], and "render" for Mathjax? Ps iPad $\endgroup$ Commented Jun 28, 2012 at 23:09
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In a commutative ring not every nonzero element has a multiplicative inverse unlike the requirement in a field that every nonzero element has a multiplicative inverse.

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