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Wikipedia says that a graded $A$-algebra is just a graded $A$-module that is also a graded ring.

Question: when one says then "finitely generated graded $A$-algebra", does one mean that every element $s$ can be written as $s=\sum^N_{i=0} a_i g_i$, where $\{g_i\}_{i=1}^N$ is a generating set, or that it can be written as $s=\sum^N_{i=0} a_i \prod_{j\in I} g_j$?

Ravi Vakil proposes in an exercise:

4.5.D. (a) Show that a graded ring $S_\bullet$ over $A$ is a finitely generated graded ring (over $A$) iff $S_\bullet$ is a finitely generated graded $A$-algebra, i.e., generated over $A=S_0$ by a finite number of homogeneous elements of positive degree.

What I don't understand is how putting the word "algebra" instead of "ring" makes the difference that the former is supposed to mean generated over homogeneous elements and the latter by "any elements". Doesn't, in general, an $A$-algebra simply mean a ring over $A$, i.e. a ring into which $A$ maps (possibly not injectively), so that an $A$-action is defined?

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  • $\begingroup$ The explanation of what a "finitely generated graded ring" means for Vakil occurs a few lines above the exercise (page 148 in the current version of the notes math.stanford.edu/~vakil/216blog/FOAGdec3014public.pdf , or page 147 according to the PDF count). It explains the difference. (For what it's worth, I have never seen this definition of "finitely generated graded ring" before.) $\endgroup$ – darij grinberg Jan 4 '15 at 21:53
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Serge Lang in his book "Algebra", defines an $A$-algebra to be any ring $B$ together with a ring homomorphism $f:A\rightarrow B$. This corroborates your interpretation.

A finitely-generated $A$-algebra is practically understood as a quotient ring of a polynomial ring in finitely many indeterminates. In other words it has the form $S=B[X_1,\dots,X_n]/I$, where $B$ is a ring, $I$ is an ideal of $B[X_1,\dots,X_n]$ and there exists a ring homomorphism $A \rightarrow B/(I \cap B)$. So far, this is independent of the grading.

Now an $A$-algebra $S$ with a $\mathbb{Z}_+$-grading (for simplicity), is a ring $S$ together with a ring homomorphism $f:A \rightarrow S$, such that $S$ can be written $S=\oplus_{i \ge 0} S_i$, and the multiplication operation of $S$, $S \times S \rightarrow S$, satisfies $S_i \times S_j \rightarrow S_{i+j}$.

I am not expert enough, but it seems to me that the phrases "graded ring $S$ over $A$" and "graded $A$-algebra $S$" indicate the same thing. This is true if we interpret the phrase "$S$ is a ring over $A$" to mean simply that we have a ring homomorphism $A \rightarrow S$, but then using Lang's definition $S$ is an $A$-algebra.

Finally, i believe that the point of the exercise you quote is to show that for a finitely-generated graded $A$-algebra, we can take the generators to be homogeneous elements.

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  • $\begingroup$ Okay, then your definition of a graded $A$-algebra $S$ is indeed a little different from Vakil's definition of a graded ring $R$ over $A$. According to Vakil, a graded ring $R$ over $A$ is a graded ring $R$ such that $R_0=A$. Also, you wrote $S = \oplus_{i>0}S_i$, where I think you meant $S = \oplus_{i\ge0}S_i$. Usually, one sets $S_+:=\oplus_{i>0}S_i$, the irrelevant ideal. $\endgroup$ – Rodrigo Jan 22 '14 at 0:18
  • $\begingroup$ Yes, i meant $i \ge 0$, thanks. So look, if we use Vakil's definition, then actually $A$ is injected into $R$. So then a graded ring over $A$ is a special case of a graded $A$-algebra. But then if you require your algebra $S$ to have the property $S_0=A$, then the two definitions coincide. So that part of the exercise is by definitions. $\endgroup$ – Manos Jan 22 '14 at 0:33
  • $\begingroup$ I suppose you meant "if you require your ring S to have the property $S_0 = A$, then the two definitions coincide." Well, after thinking about what you said in your answer I believe the point is that if you write $S$ as an $A$-algebra, i.e. $S=A[s_1,...,s_k]$, then it is more or less expected that all the $s_i$ be homogeneous. In that case, I can see how Vakil equates finitely generated $A$-algebra with being generated by homogeneous elements, although I would hope that there is a better way of arguing why it should be interpreted this way. $\endgroup$ – Rodrigo Jan 22 '14 at 0:54
  • $\begingroup$ Anyway, one should still show that given some generating set which is not homogeneous, it is possible to pick another one that is. But this follows straight from the fact that if $\{g_i\}$ are a generating set, then, after breaking each $g_i$ into its homogeneous parts $g_{ij}$ one still has a finite generating set $\{g_{ij}\}$ $\endgroup$ – Rodrigo Jan 22 '14 at 0:55
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    $\begingroup$ Regarding your 3rd comment: that's right. But the grading of the ring has to be in the non-negative integers (something that you didn't mention, but i suppose you assume). $\endgroup$ – Manos Jan 22 '14 at 1:02

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