# Misunderstanding a Theorem on Field Homomorphism

1. Dec 31, 2013

### Mandelbroth

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. Dec 31, 2013

### jgens

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.

3. Jan 1, 2014

### Mandelbroth

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.

4. Jan 1, 2014

### 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.

Last edited: Jan 1, 2014
5. Jan 1, 2014

### Mandelbroth

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!

Last edited: Jan 1, 2014
6. Jan 1, 2014

### jgens

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

7. Jan 1, 2014

### Mandelbroth

Uhhh...I don't see it. :uhh:

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

8. Jan 1, 2014

### jgens

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

9. Jan 1, 2014

### Mandelbroth

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?

10. Jan 1, 2014

### jgens

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.

11. Jan 1, 2014

### Mandelbroth

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