Let $\alpha\in\mathbb{C}$, and let $\mathbb{Z}[\alpha]$ be the smallest subring of $\mathbb{C}$ containing $\alpha$; that is, $\mathbb{Z}[\alpha]=\cap\ S$, where $S$ ranges over all those subrings of $\mathbb{C}$ containing $\alpha$. Prove that $$\mathbb{Z}[\alpha]=\{f(\alpha): f(x)\in\mathbb{Z}[x]\}$$

My attempt:

I think I know how to solve this problem but don't know how to present it as a "mathematical" proof.

Since $\mathbb{Z}[\alpha]$ is a subring of $\mathbb{C}$, it contains $0$ and $1$.

Since $\mathbb{Z}[\alpha]$ is closed under addition, it must contain all the natural numbers. It must also contain the additive inverses of the natural numbers, i.e, the negative integers. Thus $\mathbb{Z}[\alpha]$ contains $\mathbb{Z}$.

Since $\mathbb{Z}[\alpha]$ is closed under multiplication, it must contain all elements of the form $m\alpha^n$ where $m\in\mathbb{Z}$ and $n\in\mathbb{N}\cup\{0\}$.

Finally, since $\mathbb{Z}[\alpha]$ is closed under addition, it must contain all elements of the form $\sum m\alpha^n$ where $m$ and $n$ range over $\mathbb{Z}$ and $\mathbb{N}\cup\{0\}$ respectively.

How do I finish this proof and make it "mathematically presentable"?


1 Answer 1


It is easy to prove that the set you defined is a subring of $\mathbb{C}$, so $\mathbb{Z}[\alpha]$ must be contained in it by definition. On the other hand, we know that $\mathbb{Z}[\alpha]$ is closed under multiplication, sums, and contains $0$, $1$ and $\alpha$. Therefore, it must contain all polynomials in $\alpha$. Thus, $\mathbb{Z}[\alpha]$ contains the set you gave, concluding the proof.

  • $\begingroup$ Thus this generalizes to arbitrary rings $R$? i.e. $R[\alpha]=\{f(\alpha)|f\in R[x]\}$? (Where $R[\alpha]$ denotes the smallest subring of $R$ containing $\alpha$.) I was confused by the choice of $\mathbb{Z}$ there, but I might also be misunderstanding (or just missing) something. Also, how do we know $1\in\mathbb{Z}[\alpha]$? $\endgroup$ Nov 20, 2018 at 19:50
  • $\begingroup$ Correction to above: Well $1\in\mathbb{Z}[\alpha]$ above, just because $\mathbb{C}$ is a field I suppose. I was thinking more generally; But anyway, can we generalize this somehow, perhaps obtaining something similar? (Sorry if this is too much of a new question, if so you may just refer me to another answer or leave it.) $\endgroup$ Nov 20, 2018 at 19:59
  • $\begingroup$ @Christopher.L No, that is not true in general. Notice already in the example above that the subring is $\mathbb{C}$ and not $\mathbb{Z}$. However, assuming $R$ is a commutative ring, let $U$ be the smallest subring of $R$ containing $1$ (we work in unital rings). Then the same proof as above should give that the smallest subring containing $\alpha\in R$ is $U[\alpha]$ (the polynomial in $\alpha$ with coefficients in $U$). $\endgroup$ Nov 20, 2018 at 21:55

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