# What are rational integer coefficients?

I have a question about the following excerpt from Atiyah-Macdonald (page 30):

“A ring $A$ is said to be finitely generated if it is finitely generated as a $\mathbb Z$-algebra. This means that there exist finitely many elements $x_1,\dotsc,x_n$ in $A$ such that every element of $A$ can be written as a polynomial in the $x_i$ with rational integer coefficients.”

I suspect one should delete "rational" and then it says that $A$ is called finitely generated if $A = \mathbb Z [a_1, \dots a_n]$ for some $a_i \in A$, that is, every element in $A$ can be written as a polynomial in $a_i$ with integer coefficients.

If this is a typo it is not mentioned on MO but perhaps it is not and I misunderstand the definitions. If I do: What are rational integer coefficients?

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Thanks to @all of you for your helpful answers! – Rudy the Reindeer Jul 18 '12 at 14:22

Sometimes the members of $\mathbb{Z}$ are called "rational integers" to distinguish them from $\mathbb{Z}[i]$ or other rings.

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In algebra there are lots of different situations where "integers" arise, as in the context of algebraic integers. The term rational integer just distinguishes the normal integers from any other sort of algebraic integer.

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It is common to use the word "integer" to refer not only to elements of $\mathbb{Z}$, but also to a wider class of complex numbers called "algebraic integers": those numbers that are roots of a monic polynomial with integer coefficients.

In this context, "rational integer" is used for elements of $\mathbb{Z}$: it means an integer which is also a rational number.

The word "rational" also gets generalized sometimes. If we have a particular base field of interest, we call elements of that field "rational" and elements of its algebraic extensions "irrational". This reduces to the usual meaning when we take $\mathbb{Q}$ as the base field.

I see this generalization most commonly used when talking about the base field in the abstract, or when working over a particular finite field. e.g. if our base field is $\mathbb{F}_3$, then when discussing the elements of, say, $\mathbb{F}_{27}$, we would call $0,1,2$ rational, and the remaining 24 elements irrational.

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