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Let $A_0$ be a finite dimensional $k$-algebra where $k$ is an algebraically closed field of characteristic $p$ not necessarily with unit. We wish to show there is a $k$-algebra $A$ with a unit $1$ containing $A_0$ as a $k$-subalgebra of $A$ of co-dimension $1$.

Attempt

Let $\{b_1,\dots,b_n\}$ be a $k$-basis of $A_0$. Formally define $A$ to be the $k$-span of $$\{b_1,\dots,b_n\} \cup \{1\}$$ where we define $$a_0 \cdot 1 = 1 \cdot a_0 = a_0 \tag{$\forall a_0 \in A_0$}$$ and $$k \cdot 1 = \underbrace{1+\dots+1}_{k \text{ times}}$$ Then it is clear that $A_0$ is a subalgebra.

Confusion

There is a mapping of $k$ into the center of $A_0$ and so I can think of elements of $k$ in $A$, although not every element of $k$ is uniquely embedded in $A_0$. So, any $s \in k$ should be in the $k$-span of $\{b_1,\dots,b_n\}$ so whether or not my algebra $A$ is actually well-defined and whether or not $A_0$ would be of codimension $1$ seems unclear as the expression of every element should be unique. Clearly, I am losing some inuition somewhere in the definition of an algebra. Clarification would be greatly appreciated.

Follow Up Question:

I didn't use anything but finite dimensionality so I presume this holds in the much more general case of an algebra over a commutative ring $R$ where our algebra is finite dimensional over $R$?

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I think the most natural idea is to take the unitization $A^1=k\times A_0$ where addition is defined to be $(\alpha, a)+(\beta, b)=(\alpha+\beta, a+b)$ and multiplication is defined $(\alpha, a)(\beta, b)=(\alpha\beta, \alpha b+\beta a+ab)$, where the identity will be $(1,0)$, and $A_0$ is embedded as the subalgebra $\{(0,a)\mid a\in A_0\}$.

I don't find your description of the algebra in the attempt very clear, but what I am describing is much like that.

It has nothing to do with finite dimensionality of $A_0$ or algebraic closedness or characteristic of $k$.

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  • $\begingroup$ I imagine this is likely what Alperin wanted me to come up with, but it doesn't quite address my question. Do the elements of $k$ not already live in $A_0$ in some way? Doesn't that mess up the uniqueness of our linear combos? Or are we thinking about $k$ essentially being given an action on $A_0$? $\endgroup$ – Aaron Zolotor Apr 2 at 20:17
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    $\begingroup$ @AaronZolotor no, in an algebra without identity, there is not necessarily a copy of k . For example, any vector space with the zero product. The construction above furnishes at least one copy of $k$ as $k\times \{0\}$ $\endgroup$ – rschwieb Apr 2 at 23:12
  • $\begingroup$ @AaronZolotor I don't really see the necessity of talking about a basis. The Dorroh extension (the unitization I'm talking about) essentially does this: suppose I have a $k$-algebra $A_0$ and I want to lump it into a bigger ring with $k$. It would have to have things that look like $k+a$ for $k\in K$ and $a\in A_0$. What would $k+a+k'+a'$ look like? What would $(k+a)(k'+a')$ look like? After you look at that, the construction above follows naturally. $\endgroup$ – rschwieb Apr 3 at 15:19

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