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When I was reading Lang's Algebraic Number Theory, it wrote If B is integral over A and finitely generated as an A-algebra, then B is a finitely generated A-module. However, is an A-algebra automatically an A-module with an additional bilinear operation

I feel that it would be better to add the proof for this theorem given in Lang.

Proof. We may prove this by induction on the number of ring generators, and thus we may assume that $B = A[x]$ for some element $x$ integral over A. But we have already seen that our assertion is true in that case.

I am still confused with this theorem.

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  • $\begingroup$ well, it was listed as a theorem $\endgroup$ – Rikka Jan 9 '18 at 4:53
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An A-algebra is (in this context) simply a ring B which contains A as a subring.

In that situation, B can be viewed as an A-module in a hopefully obvious way.

On the other hand, if B is an A-algebra and S is a subset of B, we say that S generates B as an A-algebra if the smallest subring of B which contains A (so that it is an A-algebra) and S is B itself. Then, that B is finitely generated as an A-algebra means that there is a finite subset S of B which generates B as an A-algebra.

For example, the polynomial ring $Q[x]$ is a $Q$-algebra which is not a finitely generated $Q$-module but which is finitely generated as a $Q$-algebra: the set $S=\{x\}$ generates $Q[x]$ as a $Q$-algebra.

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    $\begingroup$ In some other contexts, one calls an A-algebra a ring B together with a fixed homomorphism of rings $f:A\to B$. Also, all the definitions I've given above are for commutative rings; if you want to deal with non-commutative algebras then the definitions have to be changed a bit. There is surely a Wikipedia page on algebras. $\endgroup$ – Mariano Suárez-Álvarez Jan 9 '18 at 5:06
  • $\begingroup$ I am sorry that I had added the proof, so it would probably be helpful. $\endgroup$ – Rikka Jan 9 '18 at 5:27
  • $\begingroup$ @MarianoSuárez-Álvarez To be more general (or symbolic, would you say :) you should consider fixed homomorphisms $f_l:A\to End(B,+)$ (resp. $f_r: A^{op}\to End(B,+)$) with some compatibility with the $B$-multiplication. This amounts to what I wrote in my (B) case. $\endgroup$ – Duchamp Gérard H. E. Jan 9 '18 at 20:14
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    $\begingroup$ A formula such as «$(\forall a\in A)(\forall b_1,b_2\in B)\Big(a(b_1b_2))=(ab_1)b_2=b_1(ab_2)\Big)$» is not used in communication beetween humans, except when one is talking about logic. $\endgroup$ – Mariano Suárez-Álvarez Jan 10 '18 at 18:03
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    $\begingroup$ Can you find one (ONE!) algebra textbook that explains what an algebra is in a form even similar to that illegible formula? As a general rule, essentially any use of the symbols $\forall$ and $\exists$ in a mathematical text except in the object language of a text on logic is misguided. $\endgroup$ – Mariano Suárez-Álvarez Jan 10 '18 at 19:31
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You have two different contexts of $A$-algebra in the literature. Both give an external definition of the situation $A\subset_{subring}B$. All depends on the context.

A) The first one (the most spread and your case) is that $B$ is a $A$-module with commutation of the multiplications i.e. for all $a\in A$ and $b_i\in B$, one has the identities (associativity w.r.t. to scaling) $$ a(b_1b_2))=(ab_1)b_2=b_1(ab_2) $$

Here, "finitely generated as a $A$ algebra" means that there exists a finite set $F\subset B$ such that the smallest $B_1$ for which
$$ A\cup F\subset_{subring}B_1\subset_{subring}B $$ is precisely $B$.

From this, you see that $B$ is finitely generated as a $A$-module (FGM) means that it exists $F=\{f_1,f_2,\cdots f_n\}\subset B$, such that for all $b\in B$ we have a decomposition $$ b=\sum_{i=1}^n a_if_i $$ with $a_i\in A$.

And $B$ is finitely generated as a $A$-algebra (FGA) means that it exists $F=\{f_1,f_2,\cdots f_n\}\subset B$, such that for all $b\in B$ we have a decomposition $$ b=\sum_{\alpha\in \mathbb{N}^F} a_\alpha F^\alpha $$ where $\alpha$ is a (weight) mapping $F\to \mathbb{N}$ i.e. $\alpha(f_i)=\alpha_i$ and $F^\alpha=f_1^{\alpha_1}\cdots f_n^{\alpha_n}$ (multiindex notation) and $a_\alpha\in A$.

So (FGM) implies (FGA).

For the converse, you need to extend $F$ with the products of powers of the $f_i$, but remaining finite. There the condition that $B$ is integral has to be used. In view of [1], for all $i\in I$, one can write $$ f_i^{d_i}=\sum_{k=0}^{d_i-1}a_k\,f_i^k $$ this proves that every $F^\alpha$ can be written as a $A$-linear combination of the $F^\beta$ with $\beta_i< d_i$ for all $i$. But those $F^\beta$ are in finite number. So $B$ is (FGM).

B) In the second one, we have $B$ is a $A$-bimodule and one has, still for all $a\in A$ and for all $b_i\in B$ (associativity w.r.t. scalings) $$ a(b_1b_2)=(ab_1)b_2\ ;\ b_1(ab_2)=(b_1a)b_2\ ;\ (b_1b_2)a=b_1(b_2a) $$ [1] wikipedia page https://en.wikipedia.org/wiki/Integral_element

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  • $\begingroup$ sorry, i am still quite confused. $\endgroup$ – Rikka Jan 9 '18 at 4:51
  • $\begingroup$ @Rikka I detailed the two situations. Apparently you are in the case of commutative rings. Am I right ? $\endgroup$ – Duchamp Gérard H. E. Jan 9 '18 at 4:59
  • $\begingroup$ Wow. You managed to turn this into a morass of symbols! :-/ $\endgroup$ – Mariano Suárez-Álvarez Jan 9 '18 at 5:00
  • $\begingroup$ @MarianoSuárez-Álvarez Wait, sometimes formal definitions are required. $\endgroup$ – Duchamp Gérard H. E. Jan 9 '18 at 5:08
  • $\begingroup$ I am supposed to be very familiar with the concept of algebras and I am not sure I understood what you wrote... $\endgroup$ – Mariano Suárez-Álvarez Jan 9 '18 at 5:10
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I'm not quite sure what you're asking. If you're asking if an $A$-algebra is automatically an $A$-algebra in a canonical way, then the answer is yes.

I'm not sure about Lang's definition, but in commutative algebra, an $A$-algebra is a commutative ring $B$ with a ring homomorphism $\phi : A\to B$. Then $B$ is naturally an $A$-module with the multiplication $m : A\times B \to B$ being given by $m(a,b) := \phi(a)b$.

It appears from the other answers that Lang uses the definition that $B$ is a ring containing $A$ as a subring. This is related to the more general definition by taking $\phi : A\to B$ to be the inclusion of $A$ in $B$.

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  • $\begingroup$ In Lang's "Algebra" he gives essentially this definition of $A$-algebra, the difference being he does not require $B$ to be commutative, so there is the requirement that the map $\phi$ sends $A$ into the center of $B.$ $\endgroup$ – Chris Leary Jan 10 '18 at 20:30
  • $\begingroup$ To follow up, I learned algebra from this book (second edition, old, like me) so the definition may have changed in subsequent editions. $\endgroup$ – Chris Leary Jan 10 '18 at 20:35
  • $\begingroup$ @ChrisLeary Thanks for the comments, I'm not going to edit, since the book in question is Lang's Algebraic Number Theory, which seems unlikely to be discussing noncommutative algebras, so I feel like the extra generality might be distracting, and because the generalization of my answer to the noncommutative case is fairly clear. $\endgroup$ – jgon Jan 10 '18 at 20:39
  • $\begingroup$ My comment was just a point of information. Your point that the question concerned algebraic number theory is well-taken. I honestly don't know the definition of $A$-algebra that Lang gave in that context. However, he did tend to be fairly consistent in his exposition. I hope my comment didn't upset you, because no offense was intended. In any event, of the answers I have seen to this question, yours is the most useful in my opinion. $\endgroup$ – Chris Leary Jan 11 '18 at 3:26
  • $\begingroup$ @ChrisLeary No offense was taken. :) In fact, I thought your comment was helpful and added useful information. I would usually edit my answer to include that information in most cases, so I wanted to respond with my reason for not editing, that's all. Hopefully I didn't come across as too brusque. $\endgroup$ – jgon Jan 11 '18 at 3:52

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