I'm given the following definitions of an $A-$algebra over a commutative ring with 1 and ring extension: an $A-$algebra $B$ is a ring $B$ together with a given ring homomorphism $\phi: A\rightarrow B$. If $A\subset B$, then $B$ is called an extension ring of $A$. A few questions arise:

  1. How does one know what the map $\phi$ is? To be specific, the author of the book I'm using states that $k[x^2]\subset k[x]$ is an (integral) extension. By the above definition it means in particular that $k[x]$ is a $k[x^2]-$algebra, so it is assumed that there is a homomorphism of rings $\phi: k[x^2]\rightarrow k[x]$. How am I supposed to understand the formula by which this $\phi$ is defined?
  2. I read somewhere that often $\phi$ is the inclusion map. What are simple examples when it is not?
  • $\begingroup$ Work through the def with simple examples. $\mathbb C$ is an $\mathbb R$-algebra. If you work through the def with this example in mind things will be more clear. $\endgroup$ – user4894 Oct 14 '16 at 20:55
  • $\begingroup$ Note that this is not the most common definition of an $A$-algebra. The usual definition requires that the image of $\phi$ be a subset of the center of $B$. By this definition, $\mathbb H$ is a $\mathbb C$-algebra. $\endgroup$ – rschwieb Oct 14 '16 at 21:13
  • $\begingroup$ @rschwieb: this definition came from a book on commutative algebra, where all rings are assumed to be commutative (and with 1). So $B$ in the question is also assumed to be commutative, I forgot to mention this. $\endgroup$ – Cary Oct 14 '16 at 21:25
  • $\begingroup$ @Cary OK, that's good to know :) Yes, that would make the two definitions match. $\endgroup$ – rschwieb Oct 14 '16 at 22:05

If $A\subset B$, then $B$ is an $A$-algebra via the inclusion map. In particular, that's what the map in your question (1) is. That is, $\phi: a\mapsto a$. As a point of language, if there is an "obvious" choice of $\phi$ (e.g. an inclusion map) and no $\phi$ is explicitly specified, then the "obvious" choice is meant. (If this strikes you as bad writing practice, I'd agree - and I've heard of examples where papers were found to contain errors because they conflated multiple "obvious" choices of $\phi$. But this is nonetheless common practice.)

For question (2), you ask when $\phi$ is not the inclusion map. Well, what if there is no inclusion map at all? For example, any ring $R$ is a $\mathbb{Z}$-algebra via the homomorphism $\phi: \mathbb{Z}\rightarrow R$ generated by $\phi(1_\mathbb{Z})=1_R$; but many rings don't contain $\mathbb{Z}$, even up to isomorphism (take $R=\mathbb{Z}/17\mathbb{Z}$, e.g.).

Even when there is an embedding, we don't need to take it as our $\phi$! For example, look at the ring $R=\mathbb{Z}[x]$. Then $R$ is an $R$-module via the identity map; but also via other maps. Consider e.g. the ring homomorphism $f: R\rightarrow R$ generated by $f(x)=x^2$. This gives $R$ a different algebra structure. I'm not an algebraist, but my understanding is that while such "unnatural" algebra structures tend to be individually not as interesting, the set of all ways of viewing $S$ as an $R$-algebra is quite an interesting object, and the "unnatural" algebra structures play important roles in it.

  • $\begingroup$ Note one important takeaway here: there will in general be many non-equivalent ways of viewing $S$ as an $R$-algebra. $\endgroup$ – Noah Schweber Oct 14 '16 at 20:56

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