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I'm wondering what ring basis is. Suppose we have a ring $R$, what basis does? For example, suppose we have a cyclotomic ring $R'$, the ring basis is defined as $\{1,\zeta,\ldots, \zeta^{n-1}\}$ where $\zeta$ is a m-th primitive root of unity and $n$ is its degree.

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    I've never heard of a "ring basis" before. I think in the example you've given, we're considering $\mathbb{Q}(\zeta_m)$ as a vector space over $\mathbb{Q}$, and you've given a basis in that sense. In general, a field extension $K$ of a field $F$ is always an $F$-vector space. – 2017-02-21
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    The example you mention only works if $n$ is prime (and if you drop the $1$), otherwise, if $n$ is composite the basis is formed of the $\zeta^k$ where $k$ is coprime to $n$. – 2017-02-22

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An integral basis for a number ring (a ring of algebraic integers) is a set of numbers such that every element in the ring is a linear combination of elements of the basis with integer coefficients.

A basis like the one you mention in the question is called a power basis.

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In some situations a ring can have a basis even without being a field. When the ring can be considered as an algebra then the notion basis makes sense. For example the ring $\Bbb Q[x,y]/(x^2+xy+y^2-x-y-2, y^3-y^2-2y+1)$ has as a basis $1, x, x^2, y, xy, x^2y$. It is not a field since with $q_1 = xy-x+1$ and $q_2 = -x^2-2*x*y+y$ we have that $q_1q_2 = (-xy-y^2+x)(x^2+x*y+y^2-x-y-2) + (2x+y)(y^3-y^2-2y+1)$ so the ring has zero divisors and is not a field.

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    You should probably add what sort of a basis this is. It is not an independent generating set for the ring as a ring for example. – 2017-02-22
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    @TobiasKildetoft: You are right, I edited my answer accordingly. – 2017-02-22
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    A couple of further things: It is important what it is an algebra over (all rings are algebras over the integers for example). The existence of a basis over whatever ring it is an algebra over is practically unrelated to whether the ring itself is a field, so I am not sure why you mention it. – 2017-02-22
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    Because a comment gave the impression that bases only existed in the framework of algebraic extensions ( so fields). But what do you do when the dimension is not finite? – 2017-02-22
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    The same as always. Modules over fields have bases, hence algebras over fields do (as this part ignores the multiplication). – 2017-02-22