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I Did Artin make a mistake regarding isomorphisms?

  1. Feb 7, 2017 #1
    I don't know if I'm losing my mind or not, but I am reading Artin's "Algebra", and in section 2.6 - isomorphisms, he writes "Corollary 2.5.9. gives us a way to verify that a homomorphism ##\varphi: G \rightarrow G'## is an isomorphism. To do so, we check that ##\mathrm{ker} \varphi = \{1\}##, which implies that ##\varphi## is injective, and also that ##\mathrm{im} \varphi = G'##, that is, ##\varphi## is surjective"

    Corollary 2.5.9: A homomorphism ##\varphi: G \rightarrow G'## is injective if and only if its kernel ##K## is the trivial subgroup ##\{1\}## of ##G##.


    Obviously Corollary 2.5.9 implies that the homomorphism is an isomorphism onto its image, but I do not see how it implies it's an isomorphism onto the co-domain.

    Counter-example: Define ##f: \mathbb{R}^+ \rightarrow \mathbb{R}^{\times}## where ##\mathbb{R}^+## is the additive group of ##\mathbb{R}## and ##\mathbb{R}^\times## is the multiplicative group of ##\mathbb{R}##. Obviously ##f## is an isomorphism onto its image (the multiplicative group of positive real numbers), but it's clearly not a surjection.

    Did Artin make a mistake or am I missing something painfully obvious?
     
  2. jcsd
  3. Feb 7, 2017 #2

    fresh_42

    Staff: Mentor

    Alright. At least in the case ##K=\{1\}##. And now you lost me.
    A monomorphism is an embedding. So if you restrict it to the image, it becomes an isomorphism as you correctly observed.
    But without this restriction, it is simply an "inclusion". Now how do you embed ##\mathbb{R}^+## in ##\mathbb{R}^\times ## exactly? I'm asking, because you cannot simply change the operation symbol, as ##0 \in \mathbb{R}^+##, but ##0 \notin \mathbb{R}^\times##.
    I get lost at "obviously".
    Very unlikely.
    More likely.
     
  4. Feb 7, 2017 #3
    ##f## is a homomorphism because ##f(x+y) = e^{x+y} = e^xe^y = f(x)f(y)##. ##f## is also clearly injective, but its image is the positive real numbers, so it isn't surjective. Thus ##f## is a homomorphism with trivial kernel, but isn't an isomorphism. Doesn't this contradict what Artin is saying? Even if my counter example is wrong, I still do not see how what Artin said is correct.

    Another counter example is ##f: \mathbb{Z}^+ \rightarrow \mathbb{Z}^+## given by ##f(x) = 2x##. ##f## is an injective homomorphism, yet isn't surjective.
     
  5. Feb 7, 2017 #4

    fresh_42

    Staff: Mentor

    Let's stay with one example. ##f=exp\, : \,\mathbb{R}^+ \rightarrow \mathbb{R}_{>0}^\times## is injective and surjective.
    ##exp\, : \,\mathbb{R}^+ \rightarrow \mathbb{R}^\times## is injective and not surjective. Everything is correct.

    But who said it was an isomorphism? Corollary 2.5.9 only states ##\varphi \textrm{ injective } \Longleftrightarrow \ker \varphi = 1##.
    The requirement ##G' = im \, \varphi## isn't made in the corollary. It is an additional condition you came up with. If this is the case, then ##G \cong G'##, but if ## im \, \varphi \subsetneq G'## then it's no isomorphism. In your example ##G'## is properly bigger than the image.
     
  6. Feb 7, 2017 #5
    We are in perfect agreement. Artin is the one that says a homomorphism ##\varphi: G \rightarrow G'## is surjective (##\mathrm{im} \varphi = G'##) if it has a trivial kernel. That's what I'm confused about! I do not see how this is true.
     
  7. Feb 7, 2017 #6
    I think I found the source of confusion. Artin wrote "To do so, we check that ##\mathrm{ker} \varphi = \{1\}##, which implies that ##\varphi## is injective, and also that ##\mathrm{im}\varphi = G'##, that is, ##\varphi## is surjective"

    I read this as him saying that the trivial kernel implies that ##\varphi## is injective as well as surjective. But upon re-reading it, it seems he means to say that after we check for the trivial kernel, we must also check that ##\mathrm{im} \varphi = G'##, which would be correct. That must mean what he meant.
     
  8. Feb 7, 2017 #7

    fresh_42

    Staff: Mentor

    That's not what is written in your quote:
    And now the extended version:
    (Fullstop)
    because this
    (pause)
    (in order to get an isomorphism)
    because
    (Fullstop)
     
  9. Feb 7, 2017 #8
    Yes, see my above post. I misinterpreted what he meant. The commas should have made it more clear, but I disregarded them. Sorry for the headache!
     
  10. Feb 7, 2017 #9

    fresh_42

    Staff: Mentor

    No problem. As a final remark: Depending on which Artin you are talking about, it might be not the original language and someone translated it. So it is rather a unlucky formulation than a real error. If it is Emil, then he was an Austrian speaking German, where a lot more commas are used to separate statements as it is the case in English. If it is his son Michael, then it's not as obvious as he is a German born American mathematician and son of an Austrian famous mathematician.
     
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