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:

  1. There is an isomorphism $\phi: F \to \phi(F) \subset K$.
  2. 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)

  1. What is the meaning of the expression $a_1 \big(x + \langle p(x) \rangle \big)$ when $a_1 \in F$?
  2. If $f(x) \in F[x]$, what can I evaluate $f$ at? Does this element need to belong to $F$?
  3. Is my restatement of the theorem correct? If both formulations are correct, how is the one given in D/F rigorous?
  • $\begingroup$ What if you evaluated $p\in R[X]$, $p(X) = X^2$ at $q\in R[X]$, $q(X) = X + 1$: $p(q(x)) = (X+1)^2$? Does it still feel strange? $\endgroup$
    – Ennar
    Mar 7 '17 at 0:36
  • $\begingroup$ While I'm at it, you probably know that $X^2+1$ has imaginary unit $i$ as root. But what is $i$? What is $\mathbb C$? Well, $\mathbb C\cong \mathbb R[X]/(X^2+1)$, and $i = X + (X^2 +1)$. $\endgroup$
    – Ennar
    Mar 7 '17 at 0:53
  • $\begingroup$ @Ennar To answer your first question: Yes, it does feel strange to evaluate $p(x) \in R[x]$ at an element not belonging to $R$. $\endgroup$
    – Sam Y.
    Mar 8 '17 at 21:19
  • $\begingroup$ Ok, I thought that might look familiar, since this is what we usually do when working with polynomials as real functions (function composition). Let's look at this abstractly. For a polynomial $p(x) = a_nx^n +\ldots + a_0$, we wonder when does $p(\alpha) = a_n\alpha^n+\ldots + a_0$ make sense, for some $\alpha$ belonging to a set $A$? First of all, we need to know how to multiply in $A$, so $\alpha^k$ makes sense, and we need to know how to add in $A$ so $\alpha^k + \alpha^l$ makes sense. Thus, we need $A$ to be ring. But, then again, we must also multiply elements of $A$ by elements of $R$. $\endgroup$
    – Ennar
    Mar 8 '17 at 22:55
  • $\begingroup$ It means that $A$ must also be an $R$-module. So, $A$ is in fact $R$-algebra. Is it sufficient as well? Yes! We conclude that we can evaluate $p\in R[X]$ at any element of $R$-algebra. In your example we have field $F$ and $F$-algebra $F[X]/I$, so we can evaluate polynomials at its elements. Hopefully, this looks less strange now. $\endgroup$
    – Ennar
    Mar 8 '17 at 22:59

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 $$ \sum_i(a_i+(p(x)))(x+(p(x)))^{i}=\sum_i a_ix^{i} + (p(x))=p(x)+(p(x))=0. $$

  • $\begingroup$ So you agree that the $p(x)$ in the theorem has coefficients from $K$ and not from $F$, i.e. the $0$ you wrote on the last line is $0$ in $K$? $\endgroup$
    – Sam Y.
    Mar 8 '17 at 21:22
  • $\begingroup$ @SamY. Yes, the $0$ in the last line is the $0$ in $K$. A field extension is nothing but a ring homomorphism $\sigma\colon k \hookrightarrow K$. If you have a polynomial $f(X)=\sum_i a_i X^{i}\in k[X]$ you may ask yourself: Does it have roots in $K$? That is to say, does the polynomial $\sum_i \sigma(a_i) X^{i} \in K[X]$ have a root in $K$, i.e. does there exists some $b\in K$ such that $\sum_i \sigma(a_i) b^{i} = 0\in K$? In the language of your first theorem, $\sigma(k) \subset K$ is an isomorphic copy of $k$ and $\sum_i \sigma(a_i)X^{i} \in \sigma(k)[X]$ has a root in $K$. $\endgroup$
    – user363120
    Mar 8 '17 at 21:40
  • $\begingroup$ @SamY. If it really bothers you that $F$ is not actually a subset of $K$ you can do the following: You have your injection $\pi \colon F\hookrightarrow K$. Set $L:=F \sqcup K\setminus \pi(F)$. Then $L$ is in bijection with the field $K$ and you can use this bijection to give $L$ a ring structure compatible with the one of $F$. Then you have a field $L$ with $F\subset L$ and the desired properties. $\endgroup$
    – user363120
    Mar 8 '17 at 21:54
  • $\begingroup$ Thanks for spelling everything out so clearly. $\endgroup$
    – Sam Y.
    Mar 8 '17 at 21:56
  • $\begingroup$ I really like the idea of making this new set $L$! But is $L$ still a field? What is the additive identity in $L$? $\endgroup$
    – Sam Y.
    Mar 8 '17 at 21:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.