Timeline for Theorem 5, Section 4.5 of Hungerford’s Algebra
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 15 at 17:20 | comment | added | SomeCallMeTim | Regarding your second question, my answer, and the second part of your first comment, I see that your definition is different from what I had in mind. I will think about it for a moment. | |
| Feb 15 at 17:19 | comment | added | SomeCallMeTim | Okay, so you mean "our defined action is not a priori a homomorphism", which is markedly different to what you wrote. The answer then is still basically the same: Having given the definition on the generators, there is only one way it can be defined on arbitrary elements, namely by linearity. Thus, this is implicitly his definition. Even ignoring this fact, the theorem is still fine, he is merely claiming that there exists a module structure satisfying this property, not necessarily defining the action there. | |
| Feb 15 at 17:13 | comment | added | user264745 | First we have to show scalar multiplication is indeed well defined. Then only we can use axioms of left module, in particular distributive axiom. I have no doubt in author’s proof. Question (1) is independent of author’s proof. | |
| Feb 15 at 17:04 | comment | added | SomeCallMeTim | Why is (left) scalar multiplication not a homomorphism? By definition, it distributes. And in the case of the tensor product, in the proof it is literally defined as $\alpha_s$, which by construction is a homomorphism. | |
| Feb 15 at 15:25 | vote | accept | user264745 | ||
| Feb 15 at 15:25 | comment | added | user264745 | Thank you so much for the answer. (1) left scalar multiplication is not homomorphism. (2) what is $s(v,w)$? Left scalar multiplication is not defined on free abelian group $F$. For clarification, here is definition of tensor product given in Hungerford’s algebra. | |
| Feb 15 at 13:37 | history | answered | SomeCallMeTim | CC BY-SA 4.0 |