Misunderstanding a Theorem on Field Homomorphism

[Mandelbroth]

Here's the theorem:

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##.

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?

 

[jgens]

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##.

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.

 

[Mandelbroth]

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.

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.

 

[jgens]

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.

 

[Mandelbroth]

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.

Oh. I've been considering this by taking the remainder before I do the products and sums. I didn't see that. Thank you.

Do you have any ideas for a shorter proof?

Happy New Year, by the way!

 

[jgens]

Do you have any ideas for a shorter proof?

There is an obvious K-algebra homomorphism K[x]→L to consider and one need only look at its kernel.

 

[Mandelbroth]

There is an obvious K-algebra homomorphism K[x]→L to consider and one need only look at its kernel.

Uhhh...I don't see it.

Could I have another hint, please? I'm sorry for this. I really should be getting this.

 

[jgens]

The homomorphism takes elements of K into elements of L via φ and takes powers of x into powers of β.

 

[Mandelbroth]

The homomorphism takes elements of K into elements of L via φ and takes powers of x into powers of β.

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?

 

[jgens]

One further simplification: No need to determine the kernel precisely. Clearly m_α is contained in the kernel, and therefore, so is the ideal it generates. This is enough to induce a map K[x]/(m_α)→L.

 

[Mandelbroth]

One further simplification: No need to determine the kernel precisely. Clearly m_α is contained in the kernel, and therefore, so is the ideal it generates. This is enough to induce a map K[x]/(m_α)→L.

Indeed. Thank you very much! You've been very helpful.