Let $k$ be a skew field. Assume that $A$ is finite $k$-algebra, i.e., ${\rm dim}_k A = [A:k] < \infty$.

Before asking I will enumerate two definitions :

Def : A $k$-algebra $A$ is $central$ if the center of $A$ is $k$.

Def : A $k$-algebra is $simple$ if only two sided-ideal is $0$ or $A$.

Question : Is there an example which is not simple but central ? Or vice versa ?

(1) Let $A=\{ X\in M_n({\bf R})|\ A$ is upper triangular and diagonal entries are same $\}$. Note that $A$ is not simple and not central. Consider ${\bf R}e_{1n}$ where $e_{1n}$ has only nontrivial entry at $(1,n)$.

(2) ${\bf H}$ is simple and central.

${\bf Reference}$ : When I study Bruer group, I found the following material : http://stacks.math.columbia.edu/download/brauer.pdf

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    $\begingroup$ Do you require $k$ to be non-commutative? Otherwise, an easy example of one of those is $\mathbb{C}$ as an $\mathbb{R}$-algebra. $\endgroup$ – Tobias Kildetoft Feb 17 '14 at 8:57
  • $\begingroup$ Yes. It need not to be non-commutative. ${\bf C}$ is simple but not central. Thank you. And maybe do you know an example of opposite case ? $\endgroup$ – HK Lee Feb 17 '14 at 8:59
  • $\begingroup$ ${\bf H}\oplus {\bf H}$ is not simple but central. $\endgroup$ – HK Lee Feb 17 '14 at 9:07
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    $\begingroup$ @HeeKwonLee Hm, well normally $\Bbb R$ in $\Bbb H\oplus \Bbb H$ would be identified with $\{(a,a)\mid a\in \Bbb R \}$, but the center of $\Bbb H\oplus \Bbb H$ is $\Bbb R\oplus \Bbb R$, so I don't think that fits the definition of "central." $\endgroup$ – rschwieb Feb 17 '14 at 13:10
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    $\begingroup$ @HeeKwonLee It's a little weird to ask for $k$ to be a skew field when algebras are conventionally defined over commutative rings, and then you are talking about $k$ being the center of an algebra and hence commutative anyway. It's not really a huge deal, but I just thought I'd throw the comment out. $\endgroup$ – rschwieb Feb 17 '14 at 13:28

You can check that the ring $T_n(k)$ of all upper triangular matrices over a field $k$ with $n>1$ is central but not simple.

To make this a complete solution for the posted question, I'd like to echo Tobais Kildetoft's excellent example already given in the comments: $\Bbb C$ is a simple noncentral $\Bbb R$-algebra.

Comment on the post (and one of its comments):

I'm not certain what you want a $k$-algebra over a noncommutative ring to be, but if you just want to talk about a ring $A$ that's a finite dimensional left vector space over $k$, we can do that. If $k$ isn't commutative, then no simple $k$-algebra (in this sense) will ever be central. So you could, for example take the $2\times 2$ matrix ring over $\Bbb H$ as a simple non-central $\Bbb H$-algebra (in this sense).

Since the center $Cen(R\oplus S)$ is $Cen(R)\oplus Cen(S)$, you won't ever be able to find an algebra whose center is a field by looking at the product of two $k$-algebras. For this reason, $\Bbb H\oplus\Bbb H$ is not central, as claimed in the comments.

  • $\begingroup$ Thank you for your introducing calculation way of center. $\endgroup$ – HK Lee Feb 18 '14 at 2:34

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