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Misunderstanding a Theorem on Field Homomorphism

  1. Dec 31, 2013 #1
    Here's the theorem:

    The author proceeds to prove the necessity of the condition, which is fairly trivial, and the sufficiency, which is less so, but there is (what I consider to be) the intuitive method of showing that ##\phi(m_\alpha)## is the minimal polynomial of ##\beta## over ##\phi(K)## (which is why I thought this way was obvious: it looked like the easiest way), and then proceeding by constructing the desired monomorphism from the isomorphism of ##K[x]/(m_\alpha)\cong \phi(K)[x]/(m_\beta)## via composition with the appropriate isomorphisms induced by the evaluation maps of ##\alpha## and ##\beta##, which the author does.

    For the sake of clarity, I'll demonstrate the full argument.

    The necessity is satisfied trivially, since ##\phi(m_\alpha)(\beta)=\tau(m_\alpha(\alpha))=\tau(0)=0##.

    Sufficiency is satisfied by the following reasoning. Again, we prove that ##m_\beta=\phi(m_\alpha)##, since ##\phi(m_\alpha)## is monic, irreducible, and has a root ##\beta##. Thus, ##(m_\alpha)## is the kernel of (what the author calls) ##q'\circ\tilde{\phi}: K[x]\to \phi(K)[x]\to\phi(K)[x]/(m_\beta)##. Thus, by the First Isomorphism Theorem, ##K[x]/(m_\alpha)\cong \phi(K)[x]/(m_\beta)##. Call this isomorphism ##\tilde{q}## and let ##i:\phi(K)(\beta)\to L## be the inclusion map, ##\tilde{E}_{\alpha}: K[x]/(m_\alpha)\to K(\alpha)## be the isomorphism from the evaluation map, and let ##\tilde{E}_{\beta}## be likewise. Then, clearly, we have the desired monomorphism by noting ##\tau=i\circ\tilde{E}_\beta\circ\tilde{q}\circ\tilde{E}_\alpha^{-1}##. (This is even more obvious if you look at a diagram of this, and I'll draw one if someone requires it.)

    The author says there is a shorter proof that does not require us to prove ##\phi(m_\alpha)=m_\beta##. I've been looking at this for about 3 hours, but I don't think I see what he means.

    Here's my best candidate for what he meant:

    We can construct a homomorphism, which I'll call ##r##, from ##K[x]/(m_\alpha)## to ##K[x]## with kernel ##(m_\alpha)## by mapping each element of ##K[x]/(m_\alpha)## to its remainder when divided by ##m_\alpha##. Clearly, ##r## is injective. We also have injective ##\tilde\phi: K[x]\to\phi(K)[x]## and ##q': \phi(K)[x]\to \phi(K)[x]/(m_\beta)##. Thus, their composition ##q'\circ\tilde{\phi}\circ r## is an injective homomorphism of rings (and, indeed, of fields) from ##K[x]/(m_\alpha)## to ##\phi(K)[x]/(m_\beta)##. We can then call this composition ##\tilde{q}##.

    Does that work?
     
  2. jcsd
  3. Dec 31, 2013 #2

    jgens

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    It is not clear (IMO) that this is actually a homomorphism. As a rule of thumb defining maps from a quotient space into the original space should give you pause.
     
  4. Jan 1, 2014 #3
    Indeed, I was troubled by this when I constructed this argument. We have the following. Let ##\overline{p}=p+(m_\alpha)##, and likewise for the other elements.

    $$r(\overline{p}+\overline{q})=p+q=r(\overline{p})+r(\overline{q}) \\ r(\overline{p}\,\overline{q})=r(\overline{pq})=pq=r(\overline{p})r( \overline{q}) \\ r(\overline{1})=1.$$

    Thus, it is a homomorphism.
     
  5. Jan 1, 2014 #4

    jgens

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    That the remainder of a sum/product is the sum/product of remainders reduced modulo the minimal polynomial (if true) requires an argument though. It is by no means obvious. The analogous statement for integers fails, for example, and I have serious doubts that it works in this case too.

    Edit: Counterexamples to your claim abound. Let r:Q[x]/(x2-2)→Q[x] be defined as before and assume it is a homomorphism. Then the following computation holds 0 = r(x2-2) = r(x2)-r(2) = r(x)2-r(2) = x2-2 and we arrive at a contradiction.
     
    Last edited: Jan 1, 2014
  6. Jan 1, 2014 #5
    Oh. I've been considering this by taking the remainder before I do the products and sums. I didn't see that. Thank you. :bugeye:

    Do you have any ideas for a shorter proof?

    Happy New Year, by the way! :biggrin:
     
    Last edited: Jan 1, 2014
  7. Jan 1, 2014 #6

    jgens

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    There is an obvious K-algebra homomorphism K[x]→L to consider and one need only look at its kernel.
     
  8. Jan 1, 2014 #7
    Uhhh...I don't see it. :uhh:

    Could I have another hint, please? I'm sorry for this. I really should be getting this.
     
  9. Jan 1, 2014 #8

    jgens

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    The homomorphism takes elements of K into elements of L via φ and takes powers of x into powers of β.
     
  10. Jan 1, 2014 #9
    Thank you. Let's see if I get this.

    We have a homomorphism ##j: K[x]\to L## that satisfies ##j|_K=\phi## and takes ##x^n## to ##\beta^n##. Its kernel is clearly ##(m_\alpha)## because ##\phi(m_\alpha)(\beta)=0##, ##m_\alpha## is irreducible, and ##K[x]## is a PID. Thus, we create ##\tilde{j}: K[x]/(m_\alpha)\to L##. Thus, we note that ##\tau=\tilde{j}\circ\tilde{E}_\alpha^{-1}##, because ##\tau|_K=j|_K=\phi## and ##\tilde{j}\circ\tilde{E}_\alpha^{-1}(\alpha)=\tilde{j}(\overline{x})=\beta##.

    Does that look right?
     
  11. Jan 1, 2014 #10

    jgens

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    One further simplification: No need to determine the kernel precisely. Clearly mα is contained in the kernel, and therefore, so is the ideal it generates. This is enough to induce a map K[x]/(mα)→L.
     
  12. Jan 1, 2014 #11
    Indeed. Thank you very much! You've been very helpful.
     
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