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I am trying to find a subring $S$ of the matrix ring $M_{p\times p}(\mathbb{F}_{q})$ over a finite field, where $p$ and $q$ are prime numbers . This subring must sastify the following three properties, which need to be proven:

1) $S$ is an $\mathbb{F}_{q}$ vector space of dimension $> 1$.

2) $S$ has the identity

3) All matrices, which are not the scalar multiple of the identity, have irreducible characteristic polynomials.

Ok, clearly, we know that the subrings has the identity of the ring. So, here the identity is the $I_{p\times p}$ matrix. I've been also reading about the irreducible matrices, their characteristic polynomials are irreducible and these matrices form a subring. Am i going in the right direction or i am completely wrong? Can anybody help me with the right idea, please? I would be glad to read your comments and remarks. Thank you in advance!

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    $\begingroup$ Perhaps I'm missing something basic but this seems to be a remarkably hard problem, in particular because of condition (3)... $\endgroup$ – DonAntonio Nov 15 '13 at 22:53
  • $\begingroup$ Yes, it's not so easy...that's why i ask for help here. But thanks for your comment. $\endgroup$ – Lullaby Nov 15 '13 at 23:02
  • $\begingroup$ Do you mean "All matrices in $\boldsymbol S\ $" in item 3)? $\endgroup$ – Matemáticos Chibchas Nov 16 '13 at 1:09
  • $\begingroup$ @DonAntonio I slightly disagree with you: I think instead that because of condition 3) the problem hardly has solution, I mean, perhaps is not that hard to show that property 3) is lost when considering sums. Summarizing, I think that the problem may be hard, but not remarkably hard (I hope not to have to eat my words later). $\endgroup$ – Matemáticos Chibchas Nov 16 '13 at 1:14
  • $\begingroup$ @Matemáticos Chibchas, yes, i mean all matrices in $S$. $\endgroup$ – Lullaby Nov 16 '13 at 9:05
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Let $K={\mathbb F}_{q^p}$, $F={\mathbb F}_{q}$, $M=M_{p \times p}(F)$. Then $K$ is a vector space of dimension $p$ over $F$, and multiplication by elements of $K$ defines linear maps of this vector space. So we can fix a basis of $K$ over $F$ and then represent the elements of $K$ as elements of $M$.

This defines an injective ring homomorphism $K \to M$ and its image $S$ is a subring of $M$ that has the required properties. The first two are clear. For the third, note that, for any element $\alpha \in K \setminus F$, we have $F(\alpha)=K$, so the minimal polynomial of $\alpha$ over $F$ (which is irreducible) has degree $p$. But the image of $\alpha$ in $S$ has the same minimal polynomial, and since this has degree $p$, it must be equal to the characteristic polynomial.

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  • $\begingroup$ Well, it seems that I am going to have to eat my words after all... $\endgroup$ – Matemáticos Chibchas Nov 17 '13 at 3:34
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    $\begingroup$ It's easy if you know the answer already, but I would imagine that it's not so easy if you have never seen it before! $\endgroup$ – Derek Holt Nov 17 '13 at 8:12
  • $\begingroup$ Thank you very much, Derek! :) $\endgroup$ – Lullaby Nov 17 '13 at 11:36

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