What is $\mathbb{Z}[a]$?

Question

Let $\alpha$ be algebraic over a field $F$. Then, $F(\alpha)$ denotes the subfield of $F$ generated by $\alpha$. This is the standard definition of $F(\alpha)$.

Under this definition, for example, one can prove that $\mathbb{Q}(\sqrt{D})=\{a+b\sqrt{D}:a,b\in\mathbb{Q}\}$ for some square free integer $D$.

Like the above example, is there a general way to define $\mathbb{Z}[a]$ for suitable arbitrary $a\in \mathbb{Z}$?

To be specific, in basic algebra texts, rings such as $\mathbb{Z}[i]$ and $\mathbb{Z}[\sqrt{-5}]$ are defined as span of $1$ and $x$. So that $\mathbb{Z}[i]$ is defined as $\{a+ib:a,b\in\mathbb{Z}\}$. However, I don't like this definition and curious to know whether there is a general definition which chooses suitable arbitrary $a$'s and form $\mathbb{Z}[a]$.

Also, I'm curious why people use the bracket $[a]$ for $\mathbb{Z}$ while people use $(a)$ for $\mathbb{Q}$. (Why $\mathbb{Z}[a]$ rather than $\mathbb{Z}(a)$ just like $\mathbb{Q}(a)$?)

Answer 1

Let $R$ be a subring of a (let's say) commutative ring with unit $A$. The intersection of subrings of $A$ containing a given subset $S$ of $A$ is again a subring of $A$ containing $S$, so that the notion of smallest subring of $A$ containing $S$ makes sense : just take the intersection of all such subrings. This smallest subring is called the subring of $A$ containing $S$. You can easily show that it is equal to the set $$\{P(s_1,\ldots,s_n)\;|\;n\in\mathbf{N}, s_1,\ldots,s_n\in S, P\in \mathbf{Z}[T_1,\ldots,T_n]\}.$$ Now if $S = R\cup T$ for a subset $T$ of $A$, this ring is noted $R[T]$ and is in fact equal to $$\{P(t_1,\ldots,t_n)\;|\;n\in\mathbf{N}, t_1,\ldots,t_n\in T, P\in R[T_1,\ldots,T_n]\}$$ and is in fact equal to the smallest sub-$R$-algebra of $A$ containing $T$. Finally, if $T = \{\alpha\}$ for an $\alpha\in A$, one notes $R[T] = R[\alpha]$, and this is equal to $$\{P(\alpha)\;|\;P\in R[x]\}.$$

Why the brackets or the parenthesis ? The notation comes from the formal variable case notation : for instance $k[T]$ for the polynomials in $T$ with coefficient in $k$ and $k(T)$ for the rational functions with coefficients in $k$ in the variable $T$. Imagine $R = \mathbf{Q}$, $A = \mathbf{R}$, and $\alpha\in A$. If $\alpha$ is a root of a non-zero from $\mathbf{Q}[T]$, then you have in fact that $\mathbf{Q}[\alpha] = \mathbf{Q}(\alpha)$ where the latter is the smallest sub-field of $\mathbf{R}$ containing $\mathbf{Q}$ and $\alpha$. But if it is not, that is, if $\alpha$ is transcendent over $\mathbf{Q}$, then $\mathbf{Q}[\alpha] \not= \mathbf{Q}(\alpha)$, and then you have an isomorphism $\mathbf{Q}(\alpha) \simeq \mathbf{Q}(T)$ inducing an isomoprhism $\mathbf{Q}[\alpha] \simeq \mathbf{Q}[T]$.

Answer 2

The definition with "span" is, at best, confusing (unless some things were explicitly agreed on a priori), and wrong at worst. By definition

$$R[\alpha]:=\left\{f(\alpha)\;:\;\;f(x)\in R[x]\right\}$$

when $\;R\;$ is any ring (commutative with unit, to make things nicer)

The brackets usually denote ring, while the round parentheses always (in the standard use) denote field of rational functions.

It is an easy and nice lemma to prove that, for example with $\;\Bbb Q\;$ and some $\;\alpha\in\Bbb C\;$ , we have that $\;\Bbb Q[\alpha]=\Bbb Q(\alpha)\iff \alpha\;$ is algebraic . Of course, we always have $\;\Bbb Q[\alpha]\subset\Bbb Q(\alpha)\;$