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

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  • $\begingroup$ 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

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