Suppose $A$ is a ring, what is A-algebra?
When I was reading Lang's Algebraic Number Theory, it wroteIf 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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$\begingroup$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.
$\endgroup$ 16 $\begingroup$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
$\endgroup$ 21 $\begingroup$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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