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I'm formalizing this useful algebraic method. In simple words: we have algebras $A,B$ (not necessarily commutative) over a field $F$. we define a function $\varphi:A \rightarrow B$ on a set of generators (of $A$ (as an $F$-module) $v_1,...,v_n$. we manually check that on these generators $\varphi$ obeys: $\varphi (a+b) = \varphi (a) + \varphi(b), \varphi (ab) = \varphi (a) \varphi (b)$ (if the generators are linearly independent, we don't have to check the addition condition). we extend it linearly by definition to an $F$-module homomorphism (this is well defined), and we amazingly also get an $F$-algebras homomorphism:

$\varphi ((\sum a_iv_i)(\sum b_jv_j)) = \varphi (\sum a_ib_jv_iv_j) = \sum \varphi(a_ib_jv_iv_j) = \sum a_ib_j \varphi (v_i) \varphi (v_j) = (\sum a_i \varphi (v_i))(\sum b_j \varphi (v_j))$

since for every $a \in F$ we get $\varphi (a) = a$.

My questions are: is this really a an F-algebras homomorphism, and is this proof valid in general? (since its my own formalization I'm not sure), and do you know any book which should contain such useful lemmas, or a graduate/bachelor's math course usually deals with such? haven't seen this on any of the textbooks I've used so far / courses I took.

Thanks, G.

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Why is $\varphi(a)=a$ for $a \in F$ ? Note that $\varphi := 0$ satisfies your requirements on additivity and multiplicativity. – tj_ Jan 26 '13 at 20:36
Well, for a linear map $\phi : A \to B$ it is clear that $\{(a,b) \in A^2 : \phi(a) \phi(b) = \phi(ab)\}$ is a linear subspace in each variable (more precisely, this defines a subspace of $A \otimes A$). Hence, it is enough to check $\phi(a) \phi(b)=\phi(ab)$ on generators of $A$ as a vector space. But your proof that $\phi$ is linear is wrong/incomplete. – Martin Brandenburg Jan 26 '13 at 23:07
@Martin: What is wrong/incomplete in the proof that $\varphi$ is linear? assuming it is linear on generators, we extend it linearly to the rest of $A$ – cruvadom Jan 27 '13 at 0:43
up vote 1 down vote accepted

Basically, what you write is true. However, there are two items that need consideration:

1) There is no guarantee that $\varphi$ is $F$-linear. So it's best to extend $\varphi$ $F$-linearly to $A$ per definition.

2) If $A,B$ have an identity, one usually requires an algebra homomorphism to preserve the identity. But $F$-linearity and multiplicativity alone are not sufficient to fulfill this requirement. For, let $A=M_2(F), B=M_3(F)$ and $\varphi: A \to B,\; X \mapsto \begin{pmatrix}X & \\ & 0\end{pmatrix}$. This map is clearly $F$-linear and multiplicative, but $\varphi(1_A) \neq 1_B$.

So, one would usually choose, say, $v_1=1_A$ and set $\varphi(1_A)=1_B$. Then, if $\varphi$ is multiplicative on the generators and extended $F$-linear to $A$, it's in fact an $F$-algebra homomorphism.

Also note that if $\varphi$ is surjective, then $\varphi(1_A)=1_B$ always holds. For, let $\varphi(x_0)=1_B$. Then $$1_B = \varphi(x_0)=\varphi(x_0\cdot 1_A)=\varphi(x_0)\varphi(1_A)=1_B \cdot \varphi(1_A)=\varphi(1_A).$$

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2) More simple $(\mathrm{id},0) : F \to F \times F$. – Martin Brandenburg Jan 26 '13 at 23:08
@tj_: 1) what do you mean by: "extend φ F-linearly to A per definition"? 2) I'm assuming $\varphi (1_A) = 1_B$, I'll add that. regardless of that, anything is wrong in proof? – cruvadom Jan 27 '13 at 0:29
1) I overlooked the sentence "we extend it linearly by definition to an F-module homomorphism". This, of course, makes $\varphi$ $F$-linear. 2) No, nothing is wrong in the proof. Also note that the property $\varphi(a)=a\;(a \in F)$ isn't used in your proof. – tj_ Jan 27 '13 at 1:19
I'm using this property in: $\sum \varphi(a_ib_jv_iv_j) = \sum a_ib_j \varphi (v_i) \varphi (v_j)$ – cruvadom Jan 27 '13 at 9:43
No: $\varphi(a_ib_jv_iv_j)\overset{(L)}{=}a_ib_j\varphi(v_iv_j)\overset{(M)}{=} a_ib_j\varphi (v_i) \varphi(v_j)$, where $(L)$ means $F$-linearity and $(M)$ multiplicativity on generators. But $\varphi(a)=a\;(a\in F)$ isn't used! – tj_ Jan 27 '13 at 11:47

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