Is a commutative A-algebra algebraic over A associative?

[coquelicot]

Let $A$ be a ring and $B$ be a commutative algebra over A·
Suppose that $B$ is generated by algebraic elements $\beta\in B$ over $A$, meaning that $\beta$ fulfils a relation of the form $P(\beta)=0$, with $P\in A[X]$.
Is $B$ necessarily associative ?

NOTE: As usual, $\beta^i$ is defined inductively by $\beta^i = \beta^{i-1}\beta$.
By commutativity, it follows that $\beta\beta^i = \beta^i \beta = \beta^{i+1}$, hence by an evident induction, $\beta^i\beta^j = \beta^{i+j}$.

I think that the answer to the question is YES because of the following reason:

We can suppose without loss of generality that $B$ is generated by a finite number of algebraic elements over $A$.

Suppose first $B=A[\beta]$ (that is, $B$ is generated by 1 element over $A$).
We have $(\beta^i\beta^j)\beta^k = \beta^{i+j+k}=\beta^i(\beta^j\beta^k)$, hence using the Cauchy product, for polynomials in $\beta$ over $A$, it follows that $B$ is associative.
If now $B$ has a finite number of algebraic generators, the result is true by an evident induction.

 

[fresh_42]

Let $A$ be a ring and $B$ be a commutative algebra over A·
Suppose that $B$ is generated by algebraic elements $\beta\in B$ over $A$, meaning that $\beta$ fulfils a relation of the form $P(\beta)=0$, with $P\in A[X]$.
Is $B$ necessarily associative ?

NOTE: As usual, $\beta^i$ is defined inductively by $\beta^i = \beta^{i-1}\beta$.
By commutativity, it follows that $\beta\beta^i = \beta^i \beta = \beta^{i+1}$, hence by an evident induction, $\beta^i\beta^j = \beta^{i+j}$.

I think that the answer to the question is YES because of the following reason:

We can suppose without loss of generality that $B$ is generated by a finite number of algebraic elements over $A$.

I don't see this. If the rest is true, you still have only a countable degree.

Suppose first $B=A[\beta]$ (that is, $B$ is generated by 1 element over $A$).
We have $(\beta^i\beta^j)\beta^k = \beta^{i+j+k}=\beta^i(\beta^j\beta^k)$, hence using the Cauchy product, for polynomials in $\beta$ over $A$, it follows that $B$ is associative.
If now $B$ has a finite number of algebraic generators, the result is true by an evident induction.

We have $\alpha \cdot (\beta_1 \cdot \beta_2) = (\alpha \cdot \beta_1)\cdot \beta_2$ and $\alpha_1 \cdot (\beta \cdot \alpha_2) = (\alpha_1 \cdot \beta)\cdot \alpha_2\,$ by definition. However, these conditions don't translate themselves automatically from $A[\beta_1]$ over $A$ to $A[\beta_1,\beta_2]=A[\beta_1][\beta_2]$ over $A[\beta_1]$ or in the next step to three. At least I don't see it immediately. I guess this is where commutativity will be needed.

I think I could construct a counterexample for non-Abelian $B$, so let me think whether this is a crucial condition or not. If the induction is correct, which I'm not sure of, then your$\text{ w.o.l.g. }$ probably does the job: $A[\beta_\iota]_{\iota \in I}=A[\beta_\iota]_{\iota \in I- \{\iota_0\}}[\beta_{\iota_0}]$ although you possibly need AC or even a transfinite induction, because the step from countable to uncountable has simply to be somewhere.

 

[coquelicot]

I don't see this. If the rest is true, you still have only a countable degree.

Yes, but any element $x$ of $B$ is a polynomial in a finite number of generators $\beta$ over $A$, so, in the process of proving that $x_1(x_2x_3) = (x_1x_2)x_3$, it suffices to consider the sub-algebra generated by suitably many generators involved in the expression of $x_1,x_2, x_3$, which are in finite number.

However, these conditions don't translate themselves automatically from $A[\beta_1]$ over $A$ to $A[\beta_1,\beta_2]=A[\beta_1][\beta_2]$ over $A[\beta_1]$ or in the next step to three. At least I don't see it immediately. I guess this is where commutativity will be needed.

Well, $A$ is a commutative ring, hence also a commutative pseudo-ring (that is, a ring without unity). Also, $A[\beta_1]$ has been shown to be a commutative pseudo-ring (since the associativity of the product has been, hopefully, shown, and the commutativity is clear). So, the same argument repeat with $A[\beta_1]$ in place of $A$ and $A[\beta_1, \beta_2]$ in place of $A[\beta_1]$ etc. I mean, the induction step could be: if $A$ is a commutative pseudo-ring and $C$ a commutative algebra over $A$ that is generated by a single algebraic element over $A$, then $C$ is a commutative pseudo ring (with respect to the algebra product).

 

[fresh_42]

I don't understand why you say "by definition" Isn't it what we are trying to prove?

The associativity with the scalars is AFAIK the $A-$algebra definition. Thus it is open, whether $A[\beta,\gamma]=A[\beta][\gamma]$ is an $A[\beta]-$algebra in this sense which is not automatically covered by induction. Or more direct: If induction works, how should we write this single induction step in terms that the algebra axiom "associativity" holds. I don't see $\alpha (\beta \gamma) = (\alpha \beta ) \gamma$ because we have the $\alpha$ part by definition for single elements $\beta$ but not how it decouples the product of two algebraic elements, i.e. stretching $\beta \gamma$ might be different from the product with a single one stretched first. Not quite sure, but here's where I would start to construct a counterexample. Although the community requirement is pretty strong, it cannot replace associativity.

Your argument about the countability is, that in all single cases we only have finitely many terms in the sums. Sounds o.k.

 

[coquelicot]

The associativity with the scalars is AFAIK the $A-$algebra definition. Thus it is open, whether $A[\beta,\gamma]=A[\beta][\gamma]$ is an $A[\beta]-$algebra in this sense which is not automatically covered by induction. Or more direct: If induction works, how should we write this single induction step in terms that the algebra axiom "associativity" holds. I don't see $\alpha (\beta \gamma) = (\alpha \beta ) \gamma$ because we have the $\alpha$ part by definition for single elements $\beta$ but not how it decouples the product of two algebraic elements, i.e. stretching $\beta \gamma$ might be different from the product with a single one stretched first. Not quite sure, but here's where I would start to construct a counterexample. Although the community requirement is pretty strong, it cannot replace associativity.

Your argument about the countability is, that in all single cases we only have finitely many terms in the sums. Sounds o.k.

You are right. Furthermore, I've realized that the assumption "$\beta$ is algebraic over $A$" has been nowhere used. Do you think, that the argument above proves at least that $A[\beta]$ is an associative sub-algebra of $B$ over $A$?

 

[fresh_42]

You are right. Furthermore, I've realized that the assumption "$\beta$ is algebraic over $A$" has been nowhere used. Do you think, that the argument above proves at least that $A[\beta]$ is an associative sub-algebra of $B$ over $A$?

Sure, since we have associativity with the elements of $A$ by definition and all other elements are polynomials in $\beta$ so it extends to them via distributivity. But this should also hold for a transcendent $\beta$ as the individual elements are all finite sums.

 

[coquelicot]

Sure, since we have associativity with the elements of $A$ by definition and all other elements are polynomials in $\beta$ so it extends to them via distributivity. But this should also hold for a transcendent $\beta$ as the individual elements are all finite sums.

Thank you so many Fresh_42.