I'm studyng basic ring theory, in a "master's degree" in math. We've studied this objects in class: $\mathbb{Z}[\sqrt[3]{2}]$ and another one similar: $\mathbb{Z}[\sqrt[3]{2}, \sqrt{6}]$.

The definition this object, K[S], is: Given a ring R, let S $\subseteq$ R. Let K be a subring of R. Then K[S] is the smallest ring that contains both R and S.

Note: My rings all have unity by definition.

The question is this: Have you heard of a similar or equal definition, and could you provide me links (e.g. wikipedia or wolfram articles, or mathSE questions). I want to know if this concept appears somewhere else or is something only known to my professor and my university. So please point me out the "name" of this K[S] (I presume is something like "Ring generated by a set"), so I can search in the internet and in books to know more about this particular object.

Note: I'm not concerned with the special cases I mentioned at the beggining, only with the name and definition of K[S].

  • 2
    $\begingroup$ This is exactly what you expected: "ring generated by a set" (technically subring for technical reasons, but for all intents and purposes, we usually consider $K[S]$ independent of the ring $R$) en.wikipedia.org/wiki/Subring#Subring_generated_by_a_set $\endgroup$ – William Stagner Sep 28 '15 at 20:00
  • $\begingroup$ @WilliamStagner I know what a subring generated by some set of the ring is, I'm not asking that, I'm asking for clarification on what K[S] is. The defiition I gave for K[S] is equivalent to $<K \cup S>$ where <D> is notation for the ring generated by D. But then why is there an assymetry in the notation? Is K[S]=S[K]? (obviously by my definition is, but the notation seems to imply they are in some relation more assymetric than just union. $\endgroup$ – Santropedro Sep 28 '15 at 20:15
  • $\begingroup$ Then I'm not quite sure what you're asking. The reason we write $K[S]$ instead of $S[K]$ is because we are intuitively adjoining $S$ to $K$ in order to form a larger ring containing $K$ as a subring. It doesn't make sense to write $S[K]$ because $S$ is just a set. $\endgroup$ – William Stagner Sep 28 '15 at 20:21
  • $\begingroup$ Yes. My definition prohibited me of writing S[K] but relaxing the definition by not asking the first parameter to be a ring, I can. The important thing is : does this object have a name? $\endgroup$ – Santropedro Sep 28 '15 at 20:41

One interpretation that may clear things up is that $K[S]$ is not only a ring, but an algebra, specifically, a $K$-algebra, specifically, the $K$-algebra generated by $S$. That clarifies the asymmetry: $\mathbb{Q}[\sqrt 3]$ is an algebra over the rational numbers, not over the square root of 3! But as a ring, it's perfectly fine to say $\mathbb{Q}[\sqrt 3]=\langle \mathbb{Q},\sqrt 3\rangle$. On the other hand, $\sqrt{3}[\mathbb{Q}]$ doesn't make much sense. At best, you could interpret it as $\mathbb{Z}[\sqrt 3][\mathbb Q]$, which correctly exhibits $\mathbb{Q}[\sqrt 3]$ as a $\mathbb{Z}[\sqrt 3]$-algebra.

It might be helpful to observe that every algebra is a quotient of a polynomial algebra, for instance, $\mathbb{Q}[\sqrt 3]$ is isomorphic to $\mathbb Q[X]/(X^2-3)$, so that the notation you're asking about is just an extension of the ordinary notation for polynomials.

  • $\begingroup$ Though this is only true if the top ring is commutative, or in general if $K \subseteq R$ makes $R$ a $K$-algebra. (For example, $\mathbb{C}[\mathbb{H}] = \mathbb{H}$ is not a $\mathbb{C}$-algebra.) $\endgroup$ – Ben Sep 28 '15 at 20:43
  • $\begingroup$ @BenDyer thanks. I never think about noncommutative rings. $\endgroup$ – Kevin Carlson Sep 28 '15 at 20:51
  • $\begingroup$ @BenDyer surely you meant "K a subset of the center of R" to be consistent with your first statement: definitions of algebras over noncommutative rings are few and far between. Your last sentence could be looked at as a counterexample to the second half of the first sentence. Regards. $\endgroup$ – rschwieb Sep 29 '15 at 2:14
  • $\begingroup$ @rschwieb Thanks, I was meaning that $(ra)\cdot(sb) = rs(a\cdot b)$ for $r,s \in K, a,b\in R$, and wasn't bothering to make any further deductions. As you say, it is equivalent to $K$ being central. But perhaps there is some value in the first version as well, since the two are no longer equivalent when we allow $R$ to be a non-unital ring, I think. $\endgroup$ – Ben Sep 30 '15 at 18:13
  • $\begingroup$ @BenDyer Yep. As one who has thought about noncommutative definitions of algebras in the past, I can appreciate that. $\endgroup$ – rschwieb Sep 30 '15 at 18:27

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