Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

4 Answers 4

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.

share|improve this answer
    
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:

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 .

share|improve this answer
    
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.

share|improve this answer
    
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?

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.