Let $K$ be a commutative ring. $M$ is a free $K$-module of rank $m$. $E(M)$ is the exterior algebra defined by $M$.
$\iota$ is a $K$-homomorphism from $M$ to $E(M)$ such that $\iota(x)=-x$. Since $\iota(x)^2=0$, $\iota$ can be extended to a unique algebra homomorphism from $E(M)$ to itself. We will use $\iota$ to denote this algebra homomorphism, and denote $\bar{a}=\iota(a)$, $a\in E(M)$.
The next part is from N.Jacobson's Basic Algebra(2nd).
I can read and understand the proof of the lemma till the last sentence above.
My question is why the image of the algebra homomorphism $f$ above has that matrix form. And actually I think I can understand the elements in position (1,1) and (2,2) of the matrix. But why the (2,1) position will be $D(a)$. And I am also confused with that $D$ becomes a $K$-endomorphism after that.
I don't know if there requires any knowledge about universal property or something like that. Actually I have learnt some about that. But I just can't figure it out.
Editted:I understand that $f$ in the original definition is a $K$-homomorphism and satisfies $f(x)^2=0$, $x\in M$. So by the universal property of $E(M)$, $f$ can be uniquely extended to an algebra homomorphism. Let's say $f'$. And $f'$ is from $E(M)$ to $M_2(E(M))$. But even though $f'$ is an algebra homomorphism, it seems that we can't get $$f'(a)=\begin{pmatrix} a & 0 \\ D(a) & \bar{a} \end{pmatrix}, a\in E(M).$$ Now the form of $f'$ can just show it is a $K$-homomorphism. At the same time, by calculation, $$f'(ab)=f'(a)f'(b)=\begin{pmatrix} ab & 0 \\ D(a)b+\bar{a}D(b) & \bar{ab} \end{pmatrix}, a,b\in E(M).$$The (2,1) position is not the expected form $D(ab)$! So again, I turn back to my origin question, why the (2,1) position of the image of $f'$ is of that form. How can one determine the value of that position?
Thank you once more for reading this!
Yes, your feeling is exactly right: it is invoking the universal property of $E(M)$.
The property is: any $K$ homomorphism from $M$ into an associative $K$ algebra $A$ such that $f(x)^2=0$ can be extended uniquely to a $K$ algebra homomorphism from all of $E(M)$ into $A$.
All of $x\mapsto x$, $x\mapsto -x$ and $x\mapsto Dx$ are extendable linear maps from $M$ into $E(M)$. They are also entries of $f$. In effect, the extension of $f$ is the matrix of extensions of its entries. The reason we can recognize the matrix is of that form is because these extensions are unique. (For clarity: the uniqueness is given by the universal property of $E(M)$, and we are not talking about the uniqueness that the lemma is striving to prove.)
It looks like Jacobson abused $D$ a little by letting it denote its extension as well as itself.