Here's the theorem:(adsbygoogle = window.adsbygoogle || []).push({});

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. Suppose that ##K(\alpha):K## is a finite simple extension, ##\alpha## has minimal polynomial ##m_\alpha\in K[x]##, and ##\phi## is a homomorphism of fields from ##K## into ##L## (which the author extends to a homomorphism from ##K[x]## to ##L[x]##, also denoting it by ##\phi##), with ##\beta\in L##. Then, there exists a (necessarily unique) field homomorphism ##\tau: K(\alpha)\to L## satisfying ##\tau(\alpha)=\beta## and ##\tau|_K=\phi## if and only if ##\phi(m_\alpha)(\beta)=0##.

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?

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# Misunderstanding a Theorem on Field Homomorphism

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