Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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

1 Answer 1

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 $$1_B = \varphi(x_0)=\varphi(x_0\cdot 1_A)=\varphi(x_0)\varphi(1_A)=1_B \cdot \varphi(1_A)=\varphi(1_A).$$

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.