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?
-
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 $\Bbb{C}$.
-
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.