Example of a reduced ring over a finite field satisfying some other conditions What is an example for:
An extension of rings $k \subset R$ where $k$ is a finite field, $R$ is a finite dimensional vector space over $k$, $R$ is reduced, and $R \neq k[r]$ for all $r \in R$.
So, initially I thought that for a prime $p$, $k = \mathbb{F}_p$ and $R = \mathbb{F}_p \times \mathbb{F}_p$ would do the trick, but note that this does not satisfy the last condition as $\mathbb{F}_p \times \mathbb{F}_p = \mathbb{F}_p[(1,0)]$ for the idempotent $(1,0) \in \mathbb{F}_p \times \mathbb{F}_p$. This is true because for any $(a,b) \in \mathbb{F}_p \times \mathbb{F}_p, (a,b) = b(1,1) + (a-b)(1,0) \in \mathbb{F}_p[(1,0)]$. This is the example our professor seems to have had in mind too, so that now that it is seen to be incorrect, I am not sure whether an example exists.
 A: Let me call admissible a reduced, finite dimensional, non monogenic  algebra (=not of the form $k[r]$) over the  finite field $k =\mathbb F_q $, $q$ a prime power.
These are the algebras you are interested in  and we shall classify them all.
If $R$ is admissible it is in particular separable (equivalently here , étale) because a finite-dimensional reduced algebra over a perfect field is separable, and so is a product $K_1\times K_2\times...  \times K_n$ of finite extension fields of $k$ .
And which products do we have to exclude because they are monogenic?
The ones of the form $R=k[X]/(F(X))$ where $F(X)$ is monic and separable.
Now write $F(X)=F_1(X).F_2(X)...F_s(X)$ where $F_i(X)$ is an  irreducible monic polynomial of degree $d_i$.
Since $F(X)$ is separable the $F_i$'s are distinct, hence generate comaximal ideals and
by the Chinese remainder theorem we have 
$$R=k [X]/(F_1(X)) \times k [X]/(F_2(X))\times ...\times k [X]/( F_s(X))\quad \quad \bigstar $$
 Now, here are the two crucial remarks which will allow us to conclude:
Remark 1
The ring $k[X]/(F_i(X))$ is the field $\mathbb F_{q^{d_i}}$, so that $\bigstar$  becomes 
$$  R=\mathbb F_{q^{d_1}}\times \mathbb F_{q^{d_2}}\times ...\times \mathbb F_{q^{d_s}}\quad \quad \bigstar \bigstar          $$ 
Remark 2
Denote  by $N(d)>0$ the number of monic irreducible polynomials of degree $d$ in $k[X]$. Then the number of   $F_i(X)$'s  of degree $d$ is at most $N(d)$, so that the number of factors in $\bigstar \bigstar $ isomorphic to  some $\mathbb F_{q^d}$ is at most $N(d)$   too. Conversely this condition is sufficient for the product to be monogenic.
Now that we have analysed the monogenic separable extensions of $k$, we can characterize the other ones, the admissible ones  you are interested in:   
Theorem
The admissible algebras are  the algebras $R= (\mathbb F_{q^{d_1}})^{e_1}\times  (\mathbb F_{q^{d_2}})^{e_2}\times...\times (\mathbb F_{q^{d_r}})^{e_r}$ in which  for at least one $i$ we have $e_i>N(d_i)$
Example
We have $N(1)=q$, the monic irreducible polynomials of degree $1$ over $\mathbb F_q$ being the $X-a, \; a\in \mathbb F_q$. So the simplest admissible algebra is $(\mathbb F_q)^{q+1}$, in line with Jyrki's example .
A: Let $R=F_p\times F_p\times \cdots \times F_p$, where there are $p+1$ factors in the product. This is clearly of dimension $p+1$ over $F_p$. Yet it is not generated by any single element. To see this, consider an arbitrary element $r=(r_0,r_1,r_2,\ldots,r_p)$, where $r_i\in F_p$ for all $i$. Because there are only $p$ elements in $F_p$, some two components of $r$ are equal by the pigeon hole principle, say $r_k=r_\ell, k<\ell$. Now if $a=(a_0,a_1,\ldots,a_p)$ is any element of the subring $F_p[r]$, we have $a_k=a_\ell$. Thus $F_p[r]$ is not all of $R$.
The generalization to other finite fields is hopefully obvious.
A: Attempt for a partial answer: Since $k\subset R$ is a finite extension, $R$ the spectrum of $R$ consists of finitely many maximal ideals $m_1,\ldots ,m_r$ and because $R$ is reduced
$
\bigcap\limits_i m_i=0.
$
Consequently $R$ is isomorphic to a product $\mathbb{F}_{q_1}\times\ldots\times\mathbb{F}_{q_r}$ of finite fields extending $\mathbb{F}_p$, where the latter is embedded diagonally into the product.
Each of the factors is generated by a primitive element $x_i$. Let $f_i$ be the minimal polynomial of $x_i$ over $\mathbb{F}_p$. Now if one can chose the $x_i$ to have pairwise distinct minimal polynomials, then the polynomial $f:=f_1\cdot\ldots\cdot f_r$  is the polynomial of smallest degree having $(x_1,\ldots ,x_r)$ as a root. Hence the element $(x_1,\ldots ,x_r)$ generates $R$ over $\mathbb{F}_p$.
In general if $x:=(x_1,\ldots ,x_r)$ is an element of $\mathbb{F}_{q_1}\times\ldots\times\mathbb{F}_{q_r}$ and $f_1,\ldots ,f_s$ are the different minimal polynomials of the elements $x_i$, then $x$ is a root of the polynomial $f:=f_1\cdot\ldots\cdot f_s$. Hence if two components $x_i\neq x_j$ possess the same minimal polynomial, the element $x$ does not generate $R$.
In Jyrki's example the possible minimal polynomials are the $p$ linear polynomials $x-c$, $c\in\mathbb{F}_p$, but we would need $p+1$ different minimal polynomials to get a generator.
A: I was reading your posts and I was wondering... what about an example of a non-monogenic and not reduced algebra over K? How do we prove that such an example must have $K$ always finite?
