I have been reading Dummit/Foote lately, and in the introductory chapter on field theory, there is the following theorem:
Theorem Let F be a field and let $p(x) \in F[x]$ be irreducible. Then there exists a field $K$ containing an isomorphic copy of $F$ in which $p(x)$ has a root.
In the proof of this theorem, $K$ is defined to be $F[x]/ \langle p(x)\rangle$, so that elements of $K$ look like $f(x) + \langle p(x) \rangle$ for some $f(x) \in F[x]$. The claim is that the element $x + \langle p(x) \rangle$ is a root of $p(x)$. My question is: If $p(x)$ belongs to $F[x]$, why can you even "evaluate" $p(x)$ at the element $x + \langle p(x) \rangle$. If $p(x) = a_nx^n + \cdots + a_1x + a_0$, evaluating at $x + \langle p(x) \rangle$ looks something like
\begin{equation*}
a_n \big(x + \langle p(x) \rangle\big)^n + \cdots + a_1 \big(x + \langle p(x) \rangle\big) + a_0
\end{equation*}
But what does $a_1 \big(x + \langle p(x) \rangle\big)$ even mean? I'm guessing that it is probably equal to $a_1x + \langle p(x) \rangle$. But as far as I can tell, there is no definition (in D/F) of what this expression means or whether it even has a meaning.
After thinking about this a little more, I have come up with the following hypothesis. Perhaps the theorem that I stated above could be stated as follows:
Theorem (modified version) Let $F$ be a field and $p(x) \in F[x]$ be irreducible. Then there exists a field $K$ such that the following are satisfied:
- There is an isomorphism $\phi: F \to \phi(F) \subset K$.
- The polynomial $\tilde{p}(x):= \phi(a_n) x^n + \cdots + \phi(a_1)x + \phi(a_0) \in K[x]$ contains a root, i.e. there exists $\alpha \in K$ with the property that $\tilde{p}(\alpha) = 0_K$.
I would feel more comfortable about using this approach. It seems strange (and to me, even not rigorous) to evaluate $p(x) \in F[x]$ at an element $\alpha \in K$ when $K$ is not (formally speaking) a super field of $F$. By using the above formulation, we don't have this problem. (Note: I know that some may take objection to my statement that $K$ is not a super field of $F$. I realize that $K$ contains an isomorphic copy of $F$, but if $F \subset K$ does not hold in the set-theoretic sense, then it seems that you run into the problem of defining what $a_1 \alpha$ means when $a_1 \in F$ and $\alpha \in K$.)
To summarize, I guess I have outlined three general questions (although I would be grateful for comments on anything in this post)
- What is the meaning of the expression $a_1 \big(x + \langle p(x) \rangle \big)$ when $a_1 \in F$?
- If $f(x) \in F[x]$, what can I evaluate $f$ at? Does this element need to belong to $F$?
- Is my restatement of the theorem correct? If both formulations are correct, how is the one given in D/F rigorous?
Best Answer
You have a canonical map $\pi \colon F \to F[x]\to F[x]/(p(x))=:K$, which is necessarily injective as $F$ is a field, i.e. we may consider $F$ as a subfield of $K$. Now the inclusion extends to $F[T] \to K[T]$, for an indeterminate $T$. In particular, the polynomial $p(T)=\sum_i a_i T^{i}$ may be considered as a polynomial with coefficients in $K$. To be precise, it looks like $\sum_i (a_i+(p(x)))T^{i}$ in $K[T]$. Now, if you evaluate this at $T=x+(p(x))$, then you see that $$ \sum_i(a_i+(p(x)))(x+(p(x)))^{i}=\sum_i a_ix^{i} + (p(x))=p(x)+(p(x))=0. $$