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I was wondering what the notation $R[a]$ really stands for, if $a\in K$, where $K$ is a ring and $R$ is a subring of $K$.

In my book they define $\mathbb{Z}[\sqrt2]=\{a+b\sqrt2|a,b \in \mathbb{Z}\}$.

So, my guess is that $R[a] = \{P(a)\mid P \in R[X]\}$. Since for $\mathbb{Z}[\sqrt2]$ this is the case, or is this just a coincidence?

hmakholm left over Monica
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Nahrm
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1 Answers1

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By definition, $R[a]$ is the smallest subring of $K$ that contains both $R$ and $a$.

As you note, if $p(x)\in R[x]$, then $p(a)\in R[a]$ necessarily. Therefore, $$\{p(a)\mid p(x)\in R[x]\} \subseteq R[a].$$ Conversely, note that $$\{p(a)\mid p(x)\in R[x]\}$$ contains $a$ (as $p(a)$ where $p(x)=x$), contains $R$ (the constant polynomials); and is a subring of $K$: every element like is $K$; it is nonempty; it is closed under differences (since $p(a)-q(a) = (p-q)(a)$); and it is closed under products (since $p(a)q(a) = (pq)(a)$). Thus, $R[a]\subseteq \{p(a)\mid p(x)\in R[x]\}$. Hence we have equality.

More generally, if $a$ is integral over $R$ (satisfies a monic polynomial with coefficients in $R$, then letting $n$ be the smallest degree of a monic polynomial that is satisfied by $a$), then $R[a] = \{r_0 + r_1+\cdots + r_{n-1}a^{n-1}\mid r_i\in R\}$.

To see this, note that clearly the right hand side is contained on the left hand side. To prove the converse inclusion, let $p(x)$ be a monic polynomial of smallest degree such that $p(a)=0$. By doing induction on the degree, we can prove in the usual way that every element $f(x)$ of $R[x]$ can be written as $f(x)=q(x)p(x) + r(x)$, where $r(x)=0$ or $\deg(r)\lt deg(p)$; the reason being that the leading coefficient of $p$ is $1$, so we can perform the long-division algorithm without problems. Thus, $f(a) = q(a)p(a)+r(a) = r(a)$, so every element of $R[a]$ can be expressed as in the right hand side.

In the case where $a=\sqrt{2}$, $R=\mathbb{Z}$, $K=\mathbb{R}$ (or $K=\mathbb{Q}(\sqrt{2})$), we have that $a$ satisfies $x^2-2$< so that is why we get $$\mathbb{Z}[\sqrt{2}] = \{ p(\sqrt{2})\mid p(x)\in\mathbb{Z}[x]\} = \{r_0+r_1\sqrt{2}\mid r_0,r_1\in\mathbb{Z}\}.$$

Arturo Magidin
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  • I don't see how can you say that this fact: $f(x)=q(x)p(x) + r(x)$ implies this other fact $f(a) = q(a)p(a)+r(a)$, since if the ring is not commutative, the evaluation of a product of polynomials is not the products of the evaluations.Did you assume commutative ring in your answer? – Santropedro Jul 14 '17 at 18:04
  • @Santropedro Yes I did (I have a long answer elsewhere about specifically the point you raise, about roots of polynomials over the quaternions); probably because the OP uses $K$ (usually reserved for fields, not for arbitrary rings) and the specific example he asks about involves commtuative rings. Moreover, in the specific paragraph in question we're talking about "integral elements", and that is almost invariably restricted to the case of commutative rings rather than arbitrary ones. – Arturo Magidin Jul 14 '17 at 19:15
  • @Santropedro Note also that in the notation $R[a]$ there is an assumption that $a$ is central. For the noncommutative case, you would use $R\langle \rangle$, or $\langle R,a\rangle$; in that case, you get $R\langle a\rangle = { p(a) \mid p(x)\in R\langle x\rangle}$, i.e., the polynomial expressions in a noncommuting variable $x$. – Arturo Magidin Jul 14 '17 at 19:21
  • Wow thank you very much, you don't imagine how useful this answer and your comment are for me!!!! I in fact saw that answer you refer to in connection with my study!! I didn't know the notation of brackets $[]$ vs angled brackets $<>$. I will investigate the new formula you gave me. – Santropedro Jul 14 '17 at 19:26
  • @Santropedro: they are usually called "free algebras". See https://en.wikipedia.org/wiki/Free_algebra Also: don't use the less-than/greater-than symbols... it's like scratching a blackboard. The LaTeX symbols are \langle and \rangle – Arturo Magidin Jul 14 '17 at 19:32
  • I've got a question ( if you want me to ask as a separate question, I totally on board with that, you decide that, you don't have to respond it also, of course) , it's if there is a characterization of the form: $R[a]={r0+r1+⋯+rn−1an−1∣ri∈R} .$ for the noncommutative case $R\langle a \rangle$. The same reasoning as before doesn't work because of my first comment. – Santropedro Jul 14 '17 at 19:37
  • Let us continue this discussion in chat. – Santropedro Jul 14 '17 at 19:54
  • @Santropedro: I don't do chat. The expressions are formal words; e.g., you would havd $r_0 + (r_{11}a + ar_{12} + r_{13}ar_{14}) + (r_{21}a^2r_{22} + r_{23}ar_{24}ar_{25} + r_{26}ar_{27}a + ar_{28}ar_{29} + \cdots)$ – Arturo Magidin Jul 14 '17 at 20:07