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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.

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up vote 4 down vote accepted

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.

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Fantastic. Thanks! – Rankeya Nov 4 '11 at 17:00

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:

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 $e_i>N(d_i)$

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 .

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Another way to see why $R$ should be a product of the form $\mathbb{F}_{q^{d_1}} \times ... \mathbb{F}_{q^{d_s}}$ is to observe that $R$ is finite dimensional over $k$, and $k$ is finite by hypothesis, thus $R$ is also finite. Hence, every prime ideal of $R$ must be maximal. Moreover, the number of elements in $Spec(R)$ is bounded above by $dim_k(R)$ (This follows by an application of Chinese Remainder). Thus, if there are $s$ maximal ideals $m_1 , ..., m_s$, since $R$ is reduced, the intersection of all the $m_i$ must be trivial, and so $R \cong R/m_1 \times ... \times R/m_s$. – Rankeya Nov 5 '11 at 13:55
((In fact up to this point, it is not necessary to assume that $k$ is finite, for it is equally true that if $k$ is infinite and $k \subset R$ a ring extension such that $R$ is finite dim over $k$, then every point of $Spec(R)$ is closed, and moreover $Spec(R)$ is a finite set. We do, however, need the assumption that $R$ is reduced to deduce $R \cong R/m_1 \times ... \times R/m_s$.)) But, each $R/m_i$ is a finite extension of $k$. So, if we assume that $k = \mathbb{F}_{q}$, then each $R/m_i = \mathbb{F}_{q^i}$. – Rankeya Nov 5 '11 at 14:01
This answer is very informative Georges. Thanks. – Rankeya Nov 5 '11 at 14:09
Dear @Rankeya, everything you say is perfectly correct. The amusing point is that I realized over lunch that my allusions to separability were superfluous and I had decided to edit my post. But now I just came back and have the pleasant surprise to see that you have spared me that work with your lucid comments. So it is my turn to thank you! – Georges Elencwajg Nov 5 '11 at 14:24
Dear @Georges: Perhaps this should be a separate question on SE altogether, but if $k \subset R$ is an extension of rings, $k$ is a field, every point of $Spec(R)$ is closed, and $Spec(R)$ is finite, then can we say anything about the dimension of $R$ over $k$ as a vector space? Or, are there examples where $R$ is infinite dimensional as a $k$ vector space? – Rankeya Nov 5 '11 at 14:48

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.

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Thanks Hagen for your response. To be honest, I was trying to show that no such example exists, and had come up with a line of reasoning similar to the one you gave. But, I am not sure how to get around the problem you point out. – Rankeya Nov 4 '11 at 16:43
So, I will not accept this as an answer yet. – Rankeya Nov 4 '11 at 16:46
Write $R=k_1\times ... \times k_r$ where $k_i$ are finite extensions of $k$. As $R=k[x]$ is equivalent to $\exists k[X]\to R$ surjective, we can see that $R=k[x]$ iff there exist $r$ pairwise distinc monic irreducible polynomials $f_1(X), ..., f_r(X)$ in $k[X]$ such that $\deg f_i(X) \mid [k_i : k]$ for all $i$. – user18119 Nov 4 '11 at 22:57
Dear @QiL, I hadn't seen your answer when I posted mine: I had gone to bed before finishing my answer! But I'm happy that we seem to have had roughly the same strategy: Les bons esprits se rencontrent :-) – Georges Elencwajg Nov 5 '11 at 10:13
Dear Georges, oui:) But I made a mistake: the condition on $\deg f_i(X)$ is $\deg f_i(X)=[k_i : k]$. You did right. – user18119 Nov 5 '11 at 13:24

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?

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I think you should address this as a separate question, if you want people to answer it. – Rankeya Nov 14 '11 at 0:13
What exactly is your question though? – Rankeya Nov 14 '11 at 0:14
To give an example for: $K$ a field and $A$ an algebra of finite dimension over $K$ that is not reduced. – Anna Nov 14 '11 at 5:13
Such an example need not have $K$ finite. Let $K$ be any field, and consider the ring $K[x,y]/(x^2,xy,y^2)$. This is finite dimensional, and you can show it is not monogenic. – Rankeya Nov 14 '11 at 5:30

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