Context. If have an algebraic element $\alpha$ over $\mathbb Q$, and I want to write $\mathbb Z[\alpha]:=\{a+\alpha b,\ a,b\in \mathbb Z\}$ as a quotient ring of the form $\mathbb Z[X]/I$.

Is the following approach correct?

Let $\pi$ be an irreducible of $\mathbb Z[X]$ such that $\pi(\alpha)=0$.

Then let's consider the function

$$ \begin{matrix}\varphi\colon& \mathbb Z[X] & \to & \mathbb Z[\alpha] \\ &P& \mapsto& P(\alpha).\end{matrix}$$

The function $\varphi$ is a surjective ring morphism.

Plus, if $\varphi(P)=0$, then let's do the euclidean division (in $\mathbb Q[X]$) of $P$ by $\pi$:

$$P=Q\pi + R.$$

So we have $Q(\alpha)\pi(\alpha)+R(\alpha)=0$, so $R(\alpha)=0$ since $\pi(\alpha)=0$.

So $P\in (\pi)$ where $(\pi)=\pi\mathbb Z[X]$ the ideal generated by $\pi$.

Reciprocally, if $P\in (\pi)$ we obviously have $P\in \mathrm{ker}(\varphi)$.


$$\mathbb Z[X]/(\pi)\simeq \mathbb Z[\alpha].$$


Thanks to a comment, I should assume certain conditions on $\alpha$ which would assure that $\mathbb Z[\alpha]$ is a ring. It seems that $\alpha$ is algebraic of degree $2$ is sufficient, so I will be assuming this.

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    $\begingroup$ Yes, it is correct. Not too detailed, but for anyone knowing the basics of ring theory and, in particular, of rings of polynomials over integral domains, more than enough. Nice.+1 $\endgroup$ – DonAntonio May 12 '17 at 16:31
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    $\begingroup$ In fact $\mathbb{Z}[X]$ is not an euclidean domain. What do you mean by euclidean division in this context? $\endgroup$ – Severin Schraven May 12 '17 at 16:34
  • $\begingroup$ @SeverinSchraven I think I can look at the euclidean division in $\mathbb Q[X]$? $\endgroup$ – E. Joseph May 12 '17 at 16:36
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    $\begingroup$ @E.Joseph $\mathbb{Z}[\alpha] = \{\sum_{n=0}^d c_n \alpha^n, c_n \in \mathbb{Z}, d \in \mathbb{N}\}$. If $\alpha$ is algebraic then $\sum_{n=0}^m b_n \alpha^n = 0$ so that $\alpha^m = -\sum_{n=0}^{m-1} \frac{b_n}{b_m} \alpha^n$. If $ \frac{b_n}{b_m} \in \mathbb{Z}$ then $$\mathbb{Z}[\alpha] = \{\sum_{n=0}^m c_n \alpha^n, c_n \in \mathbb{Z}\}$$ (try with $\alpha= \frac{\sqrt{2}}{2}$ where this doesn't work) $\endgroup$ – reuns May 12 '17 at 16:59
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    $\begingroup$ Note: ${\Bbb Z}[\alpha]$ is always a ring; it is $\{a + b\alpha \mid a,b\in{\Bbb Z}\}$ if and only if $\alpha$ is algebraic of degree at most $2$. $\endgroup$ – Magdiragdag May 12 '17 at 18:36

The way you have defined it, it is not even a ring

  • $\begingroup$ $\mathbb Z[\alpha]$ is not a ring? $\endgroup$ – E. Joseph May 12 '17 at 16:37
  • $\begingroup$ not the way u have defined it $\endgroup$ – user379195 May 12 '17 at 16:37
  • $\begingroup$ $\{a+\alpha b,\ a,b\in \mathbb Z\}$ is a ring iff $\alpha$ is a root of a monic polynomial with integer coefficients. $\endgroup$ – lhf May 12 '17 at 16:42
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    $\begingroup$ a monic polynomial with degree 2 $\endgroup$ – user379195 May 12 '17 at 16:46
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    $\begingroup$ I didn't even noticed this the first time: Soumik is absolutely correct. The way it is defined in the question, $\;\Bbb Z[\alpha]\;$ is a ring only if $\;\alpha\;$ is an alg. number of degree two. The general case for algebraic numbers of degree higher is different. Not so much messier, but more general. $\endgroup$ – DonAntonio May 12 '17 at 16:51

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