Background:

In the context of Atiyah-Macdonald (so all rings are considered commutative and unital) an $R$-algebra, $A$, is defined to be a ring $A$ with a homomorphism $f\colon R \to A$. As mentioned in the text, the point is that the homomorphism induces a compatible $R$-mod structure on the ring $A$.

First, I believe that this definition actually defines an associative algebra but that is not my point here. I have also seen the alternate definition of an $R$-algebra that does not mention the homomorphism $f$, but simply requires that $A$ have both a ring structure and a compatible $R$-mod structure. That is fine with me.

Question: Out of the three objects rings, modules, and algebras, the last is the one that I have the least experience with. Something that is particularly confusing to me is the notion of an algebra being generated by a set of elements. The way this "generating" process is defined is not intuitive to me. The notions of an ideal being generated by a set of elements or a module generated by a set of elements seems to be the intuitive process to me, just analogues of 'span' from linear algebra - taking linear combinations with coefficients from the appropriate place. This is what I naturally think of anytime an algebraic object is said to be generated by a set of elements. However, Atiyah-Macdonald defines the $R$-algebra $A$ to be generated by a set $B$ if every element of $A$ can be expressed as a polynomial in the elements from $B$ and coefficients in $f(R)$ (if the definition not mentioning the homomorphism $f$ is used, define $f(r) = r\cdot 1$, then use $f(R)$ as stated). It seems to mean that, if you wanted a explicit operational description of "algebra generated by", it would be something like:

The $R$-algebra $A$ is generated by $\{x_0\}$ if for all $a \in A$, $$a = \sum_{i=0}^n r_i x_0^n$$ for some $n \in \mathbb{N}$ and $r_i \in f(R)$.

My question is why is the notion of 'generated' for algebras so much different than the notion of that algebra being generated as a module? For example, $k[x]$ requires infinitely many generators as a $k$-mod but requires only one as a $k$-algebra? I suspect that my lack of experience with algebras, and how they are used, is why it is not intuitive to me. The notion for algebras seems to be connected to integral dependence?

• The notion of generation is pretty much the same for groups, modules and algebras. $x_1,\ldots,x_n$ generate a group if the only subgroup containing them is the whole group. $x_1,\ldots,x_n$ generate a module if the only submodule containing them is the whole module. $x_1,\ldots,x_n$ generate an algebra if the only subalgeba containing them is the whole algebra. – Lord Shark the Unknown Jan 27 '18 at 22:17
• "I believe that this definition actually defines an associative algebra" Just as you observed that in the context of that book rings are commutative, well, algebras are associative. If that were your point, it would be a silly one! – Mariano Suárez-Álvarez Jan 27 '18 at 22:34
• The definition A-M uses and the alternative one you mentioned are exactly the same. If you have an algebra on the sense you mean, defining f(x)=x.1 (where on the right the dot denotes the module structure) you get an algebra in the book's sense. – Mariano Suárez-Álvarez Jan 27 '18 at 22:37
• Right, I just mentioned the bit about associative because I tagged this 'algebras' and was referring to the object as an algebra, so I was just letting the more informed reader know that I am aware I am dealing with a smaller class of objects, associative algebras, and to the second comment, yes, I also mentioned I have seen the proof of equivalence and I even mention defining the morphism $f(x) = x\cdot 1$ if the alternate definition is used. – Prince M Jan 27 '18 at 22:43

You're thinking about "generating" wrong; it doesn't have anything particularly to do with linear combinations. A subset $S$ of an algebraic structure $A$ "generates" an element $a\in A$ if you can produce $a$ by starting with the elements of $S$ and repeatedly using the operations of $A$.

In the case that $A$ is an $R$-module, the operations you have are addition and scalar multiplication by any element of $R$. So for instance, if $s_1,s_2,\dots$ are elements of $S$ and $r_1,r_2,\dots$ are elements of $R$, some elements you can generate from $S$ are: $$s_1+s_2$$ $$r_4(r_1r_2(s_1+r_3s_2)+s_3)$$ $$r_3(s_1+s_2+r_1s_3+r_2s_4)+r_4r_5s_5+s_1$$ At a first glance, these might look a lot more complicated than just linear combinations! However, since scalar multiplication is associative and distributes over addition, you can always reduce any such combination to a linear combination. For instance, the second element above is equal to the linear combination $$r_5s_1+r_6s_2+r_4s_3$$ where $r_5=r_4r_1r_2$ and $r_6=r_4r_1r_2r_3$.

In the case that $A$ is an $R$-algebra, you have one additional operation: besides addition and multiplication by elements of $R$, you can also multiply elements of $A$ together. So some elements which you can generate from $S$ are: $$s_1s_2$$ $$r_1(s_1+s_2)(s_3+r_2s_4)+s_4^3$$ $$(r_1s_1+r_2s_2)^2s_3+s_1$$ This time we can't just reduce everything to linear combinations of the $s_i$, because we might be multiplying multiple $s_i$ together, as in $s_1s_2$. However, we still can use the ring axioms to reduce everything down to a linear combination of products of the $s_i$. For instance, the second element above is equal to $$r_1s_1s_3+r_3s_1s_4+r_1s_2s_3+r_3s_2s_4+s_4^3$$ where $r_3=r_1r_2$. This is a linear combination of the products $s_1s_3, s_1s_4,s_2s_3,s_2s_4,$ and $s_4^3$. Another name for "linear combination of products" is "polynomial". Indeed, the expression above is just $f(s_1,s_2,s_3,s_4)$ where $f$ is $$f(x,y,z,w)=r_1xz+r_3xw+r_1yz+r_3yw+w^3,$$ a polynomial in four variables with coefficients in $R$. So in general, the elements of the algebra generated by $S$ are all elements of $A$ which can be written as polynomials (in multiple variables) with coefficients in $R$ evaluated at elements of $S$.

• Wow, you actually basically cleared it up for me in the first paragraph. "using the operations of $A$". So in the $k[x]$ example we can get the $x^2$ with the ring structure, but just using the module structure we need $x^2$ as a separate generator. Of course, I believe we have to define the empty product in order to generate the ground field for this example to work but thats a nonissue. I don't know why this wasn't more obvious to me? I think part of my issue was that I was also comparing to the definition of an ideal – Prince M Jan 27 '18 at 22:48
• "generated by", where we do have the ring structure but it is actually defined more like the 'linear combination' method I mention. However, you could define it more like how it is for an algebra, it just wouldn't matter since you can build higher powers anyways by choosing the generating as the ring element you are multiplying by. – Prince M Jan 27 '18 at 22:49
• Great answer by the way, still not sure what the heck I was thinking but this is exactly what I needed to clear little road block up. – Prince M Jan 27 '18 at 22:54
• Yeah, I glossed over the fact that "products" include the empty product which gives you the unit element. – Eric Wofsey Jan 27 '18 at 23:01
• No worries, even A-M doesn't mention it (that I have noticed). Thanks again! – Prince M Jan 27 '18 at 23:04

I'll take a couple of examples that will make things clearer – I hpe.

Consider the fields $\mathbf Q$ and $\mathbf Q(i)\subset \mathbf C$. It is finitely generated $\mathbf Q$-algebra, as it is the same as $\mathbf Q[i]$, and we have a surjective $\mathbf Q$-algebras homorphism: \begin{align} \mathbf Q[X]&\longrightarrow \mathbf Q[i],\\ X&\longmapsto i. \end{align} Its kernel is the minimal polynomial of $i$ over $\mathbf Q$, so $\mathbf Q(i)=\mathbf Q[i]\simeq\mathbf Q[X]/(X^2+1)$. It is finitely generated as a $\mathbf Q$-vector space (of (dimension $2$) since, as you know, all powers of $i$ are $1,i,-1$ or $-i$.

On the other hand, you may consider the $\mathbf Q$-algebra $\mathbf Q[\pi]$. It is finitely generated and you may define a similar surjective homomorphism: \begin{align} \mathbf Q[X]&\longrightarrow \mathbf Q[\pi],\\ X&\longmapsto \pi. \end{align} This one is injective, since $\pi$ is not an algebraic number. As a consequence $\mathbf Q[\pi]$ is isomorphic to the polynomial ring $\mathbf Q[X]$, so that, as a $\mathbf Q$-vector space, it is infinite-dimensional.