In Kemper's book 'A Course on Commutative Algebra' there is a question (2.5 specifically) where one has to show that given a field $K$, a sub-algebra $A\subseteq K[x_1, ..., x_n]$, if we have a $S\subseteq A$ which generates $A$, as an algebra, then $S$ is separating. I'm not actually interested in how to show this as it is straightforward once the definition of 'generated' is clear, but it rather confused me. I figured that $S$ generating $A$ as an algebra means that every $f\in A$ can be written as $\sum_{i = 1}^n g_i h_i$, where $g_i\in S$ and $h_i\in K[x_1, ..., x_n]$, just like one would generate a module. This did not yield a proof after some work, so I figured my definition must be wrong but Googling didn't provide any alternatives. I then looked at the solution. Kemper says this:

"That $S$ generates $A$ means that for every element $f\in A$ there exist finitely many elements $f_1, ..., f_m \in S$ and a polynomial $F\in K[T_1, ..., T_m]$ in $m$ indeterminates such that $f = F(f_1, ..., f_m)$.

I suspect my mistake is simple, but I currently can't see the forest for the trees. Is how I generate elements based on such an $S$ incorrect? It seems like the natural way to generate elements based on how one would generate elements in modules to me.

EDIT: After a little more searching, I discovered I should have been more thorough. The question had been asked before: Definition of a finitely generated $k$ - algebra

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    $\begingroup$ the definition of S generating A as a module is the definition given by you , for an algebra is the definition given in the statement mentioned above. The definition of "generating" changes when moving from a module to an algebra. $\endgroup$ – Chirantan Chowdhury Mar 13 '17 at 18:12
  • $\begingroup$ So what would the definition of "generating" be over an algebra? I feel very silly but I seemingly can't find it anywhere. $\endgroup$ – rwmak Mar 13 '17 at 18:18
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    $\begingroup$ Understand in this manner by this statement x,y generate k[x,y] as an algebra over k $\endgroup$ – Chirantan Chowdhury Mar 13 '17 at 18:20
  • $\begingroup$ So would it then be elements of the form $\sum_{i=1}^n a_i x_i^n + \sum_{j=1}^m b_j y_j^m + c$ for $a_i, b_i\, c\in k$ and $m, n$ arbitrary positive integers? That does coincide with the definition Kemper uses. $\endgroup$ – rwmak Mar 13 '17 at 18:24

In general, if a "widget" is a kind of structure, $A$ is a widget, and $S\subseteq A$ is a subset, then $S$ generates $A$ if there is no proper subwidget of $A$ that contains $S$. So in this case, $S$ generates $A$ as a $K$-algebra if there is no proper $K$-subalgebra of $A$ that contains $S$.

What does this mean concretely? Well, if a $K$-subalgebra of $A$ contains $S$, it must contain all $K$-linear combinations of products of elements of $S$. Conversely, it is easy to see that the set of such $K$-linear combinations is itself a $K$-subalgebra of $A$, and so since it cannot be a proper subalgebra it must be all of $A$. So $S$ generates $A$ as an algebra iff every element of $A$ can be written as a $K$-linear combination of products of elements of $S$. Equivalently, such a linear combination of products can be thought of as a polynomial with coefficients in $K$, where you have substituted certain elements of $S$ for the variables, which yields Kemper's criterion.

  • $\begingroup$ Thanks for the general explanation, it really helped me link 'generating' as a concept. $\endgroup$ – rwmak Mar 14 '17 at 18:45

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