# A basic theorem on field isomorphisms

I'm reading A Book Of Abstract Algebra by Charles C. Pinter. On page 314 is the following theorem:

Let $h:F_1\to F_2$ be an isomorphism, and let $p(x)$ be irredicible in $F_1[x]$. Suppose $a$ is a root of $p(x)$, and $b$ is a root of $h(p(x))$. Then $h$ can be extended to an isomorphism $\bar h:F_1(a)\to F_2(b)$, and $\bar h(a)=b$.

By $h(p(x))$ the author is referring to the obvious extension of $h$ to $F_1[x]$ (applying $h$ to each coefficient).

Noting that each element of $F_1(a)$ can be expressed in a unique fashion as $\sum c_i a^i$, the author defines $$\bar h(\sum c_i a^i)=\sum h(c_i) b^i$$

This seems to me to be well-defined, always a homomorphism, and bijective as long as $a$ has the same degree over $F_1$ as $b$ has over $F_2$. The author's requirement that $b$ be a root of $h(p(x))$ seems unnecessarily strong.

Question: Is this theorem true so long as $a$ has the same degree over $F_1$ as $b$ has over $F_2$?

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No, the requirement that $b$ be a root of $(h(p))(x)$ cannot be dropped. Indeed,

$$0= h(0) = h(p(a)) = (h(p))(h(a)) = (h(p))(b)$$

shows that $b$ is a root of $h(p)$.

Vaguely speaking, a map between two objects must preserve existing relations. The condition that the relation $p(a)=0$ be preserved by $h$ is precisely that $h(a)$ be a root of $h(p)$.

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How do you obtain $...=(h(p))(h(a))$ ? – Jack M Nov 26 '13 at 16:22
@JackM If $p(x) = \sum c_n x^n$ then by definition $h(p)(x) = \sum h(c_n) x^n$. Thus $h(p(a)) = h(\sum c_n a^n) = \sum h(c_n) h(a)^n = (h(p))(h(a))$. – Bruno Joyal Nov 26 '13 at 16:34
Shouldn't that technically be $\bar h$, since you're applying it to things outside of $F_1$? – Jack M Nov 26 '13 at 16:37
@JackM More or less. We are trying to construct $\bar h$, so it wouldn't make sense to use it in the construction. The fact is that the notation $h(a)$ makes sense regardless of anything else: it's just the element $\sum c_n a^n$ of $F_1(a)$. – Bruno Joyal Nov 26 '13 at 16:41

Here is a simple example that your condition is not enough:

Take $F_1 = F_2 = \mathbb{Q}$, the isomorphism $f$ being the identity. Take $p(x) = x^2-2$ and take $a=\sqrt{2}$, $b=\sqrt{3}$. Then suppose $f=\mathrm{id}$ can be extended to $\bar{f}\colon\mathbb{Q}(\sqrt{2})\to\mathbb{Q}(\sqrt{3})$ with $\bar{f}(\sqrt{2}) = \sqrt{3}$. It then follows

$$2 = f(2) = f(\sqrt{2}\sqrt{2}) = f(\sqrt{2})^2 = \sqrt{3}^2 = 3$$

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This is a convincing example, but I'm having trouble seeing at which point, in the general case, the proof that $\bar h$ is an isomorphism breaks down. I don't see which step fails to apply if $b$ isn't a root of $h(p)$. – Jack M Nov 26 '13 at 16:39