Algebras $A_i$ generated by a single element over an infinite field, does $A_1 \times \cdots \times A_r$ has the same property? Let $K$ be an infinite field and $A_1, \ldots, A_r$ finite dimensional algebras over $K$ and such that $\forall i = 1, \ldots, r \ \exists x_i \in A_i :  A_i = K[x_i]$. (I think we say that $A$ is finitely generated ?)
I would like to know if $A_1 \times \cdots \times A_r$ has the same property.
As $K[x]$ is finite dimensional, I know that we have a minimal polynomial $\pi_x$ and $\{1,x,\dots,x^{d-1}\}$ is a basis for $K[x]$ where $d$ is the dimensio but I am stuck to prove that the product has the same property. I don't have any course on algebra, it's just out of curiosity.
How can I proceed, please ? 
 A: If we follow Bourbaki in calling a  $K$-algebra $A=K[x]$ generated by a single element monogeneous then, the answer to your question is "yes":
Proposition: the product of two  finite monogeneous algebras over an infinite field $K$ is monogeneous. 
The two ingredients of the proof are:
1) A finite monogeneous algebra $A=K[x]$ is an algebra of the form $A=K[T]/\langle f(T)\rangle$, where $f(T)$ is the minimal polynomial of $x$ over $K$.
Note carefully that $f(T)$ is monic but not irreducible if $A$ is not a field.
 2) If $f(T)$ is the product $f(T)=g(T)\cdot  h(T)$ of two relatively prime polynomials we have, by the Chinese remainder theorem  an isomorphism of $K$-algebras $$A=K[T]/\langle f(T)\rangle =K[T]/\langle g(T)\rangle\times K[T]/\langle h(T)\rangle$$  The proof is now easy:
If the given two monogeneous algebras are $B=K[T]/\langle g(T)\rangle$ and $C= K[T]/\langle h(T)\rangle$, find $q\in K$ such that $g(T)$ and $h(T+q)$ are relatively prime: this can be done because the field $K$ is assumed infinite .
Since $$C=K[T]/\langle h(T)\rangle=K[T+q]/\langle h(T+q)\rangle=K[T]/\langle h(T+q)\rangle$$ we deduce $$B\times C=K[T]/\langle g(T)\rangle \times K[T]/\langle h(T+q)\rangle=K[T]/\langle g(T)\cdot h(T+q)\rangle  $$ and we have proved that $B\times C$ is indeed monogeneous.   
Remark
The Proposition is false for every finite field $K$ with $q$ elements.
For example  the product algebra $A=K^N$ is not monogeneous when  $N\gt q$ because it has $N$ algebra morphisms to $ K$ (the $N$ projections) whereas any algebra $K[T]/\langle f(T)\rangle $ has as many algebra moprphisms to $K$ as $f(T)$ has different roots in $K$ and thus at most $q$.
A: What you have written down says that $A_i$ is generated by a single element. This is much stronger than being finitely generated! The product of finitely many finitely generated $k$-algebras is again a finitely generated $k$-algebra.
In fact, if $A$ is generated by $a_1,\dotsc,a_n$ (as an algebra!) and $B$ is generated by $b_1,\dotsc,b_m$ (as an algebra!) , then one checks that $A \times B$ is generated by $(a_1,0),\dotsc,(a_n,0),(0,b_1),\dotsc,(0,b_m)$ (as an algebra!) .
