# Given a $k$ algebra $A_0$, find another $k$ algebra $A$ with identity such that the extension $A_0\subseteq A$ has dimension $1$.

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.

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 $$R$$?

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 $$\{(0,a)\mid a\in A_0\}$$.
It has nothing to do with finite dimensionality of $$A_0$$ or algebraic closedness or characteristic of $$k$$.
• 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 $A_0$? – Aaron Zolotor Apr 2 at 20:17
• @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 $k\times \{0\}$ – rschwieb Apr 2 at 23:12
• @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. – rschwieb Apr 3 at 15:19