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?
Is there any difference between the definition of a commutative ring and field?
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13No, all non-zero element in a field must be invertible, which need not hold in an arbitrary ring. – 2012-06-28
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0So this is the property that makes those two sets different – 2012-06-28
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6The other difference is that in a field $1$ must be different from $0$. – 2012-06-28
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5Why did you think they were the same? – 2012-06-28
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0I 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 – 2012-06-28
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0@ChrisEagle is 1 equal with 0 in commutative ring? – 2013-04-10
2 Answers
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:
The ring of polynomials in one indeterminate over $\Bbb{Q}, \Bbb{R}$, $\Bbb{C}$, $\Bbb{F}_{11}$, $\Bbb{Q}(\sqrt{2},\sqrt{3})$ or $\Bbb{Z}$.
The quotient ring $\Bbb{Z}/6\Bbb{Z}$
$\Bbb{Z}[\zeta_n]$ - 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)
The direct sum of rings $\Bbb{R} \oplus \Bbb{R}$ that also has the additional structure of being a 2-dimensional $\Bbb{R}$ - algebra.
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 $X$.
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.
I believe when $G$ is a cyclic group, the endomorphism ring $\textrm{End}(G)$ is an example of a commutative ring.
Examples of Fields:
$\Bbb{F}_{2^5}$
$\Bbb{Q}(\zeta_n)$
$\Bbb{R}$
$\Bbb{C}$
The fraction field of an integral domain
More generally given an algebraic extension $E/F$, for any $\alpha \in E$ we have $F(\alpha)$ being a field.
The algebraic closure $\overline{\Bbb{Q}}$ of $\Bbb{Q}$ in $\Bbb{C}$.
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3Nitpick: in #5, if $X$ has only one point, then $C(X)$ is a field. – 2012-06-28
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1The first line does not compile for me. Is it just me (+ 1:30am)? – 2012-06-28
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1Fine on the iOS – 2012-06-28
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0@AsafKaragila I'm on chrome with Ubuntu 11.04 and it loads for me. – 2012-06-28
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0@TheChaz Are you on MAC? – 2012-06-28
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0You clearly misunderstood me. I meant that the syntax of the first line does not compile for me. Aren't you missing a comma, a period, a semicolon or a colon after "No"? :-) – 2012-06-28
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0@AsafKaragila Sorry, what do you mean by "the syntax of the first line does not compile for me"? Are you referring to LaTex not rendering or something? – 2012-06-28
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0No, the English. – 2012-06-28
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0@AsafKaragila Is it better now? Sorry this is just copy and paste from the way I speak, when I type up stuff I imagine I'm talking to someone; sometimes it just carries over without making sense... – 2012-06-28
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0Well, if you ever intend to write articles, notes or books to be read by other people you need to get used to not doing that, or at least grok the punctuation rules and how they are translated from actual speech in terms of pauses (and where they should come in the text, and when speaking). – 2012-06-28
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1How about we use "parse" for engrish [sic], and "render" for Mathjax? Ps iPad – 2012-06-28
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.