I'm starting to study Field Theory by myself, the books don't say explicitly what a polynomial is, I mean, what the $x$ of $f(x)$ in $F[x]$ is? $x\in F$? When I take $f(\alpha)$ am I taking the element of F: $f(\alpha) = a_0 + a_1\alpha +a_2\alpha^2...$ where + is the sum of the field?

I hope you understand what I mean.

Sorry about my question, I know is a really trivial question, but I'm a really beginner.



If $f(x) \in F[x]$, then $f(x)$ is a finite sum $a_0+a_1x+a_2x^2+\cdots+a_nx^n$, where all the $a_i \in F$ and the $x$ is just a "formal" symbol. So, no, the $x$ is not in $F$, but is just a symbol, often called an indeterminate.

Yes, if $\alpha \in F$, then by $f(\alpha)$ we just mean the polynomial obtained by replacing each occurence of $x$ by $\alpha$. Think of the well-known $\mathbb{R}[x]$ (you've been working in this ring since high school), and the polyomial $f(x)=2x+4 \in \mathbb{R}[x])$. Then, for example, if $2 \in \mathbb{R}$ (this is your $\alpha$), $f(2)=8 \in \mathbb{R}$.

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    $\begingroup$ In order to answer better next time, I'd be grateful for the reason for the downvote. $\endgroup$ – Fredrik Meyer Oct 18 '12 at 19:13
  • $\begingroup$ so $f(\alpha) \in F$? $\endgroup$ – user42912 Oct 18 '12 at 19:17
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    $\begingroup$ I didn't downvote it myself, but I think your explanation is too brief and cryptic to help a confused beginner. Not that there anything technically wrong with it. $\endgroup$ – Harald Hanche-Olsen Oct 18 '12 at 19:19
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    $\begingroup$ @Fredrik Your answer does not appear to address the essence of the matter, viz. what is a formal polynomial? For the dangers of speaking about such objects informally, see Hankel's scathing critique of Cauchy's definition of $\Bbb C,$ which concludes with "I do not think I'm exaggerating in calling this an unintelligible play on words, ill-becoming of mathematics, which is proud and rightly proud of the clarity and evidence of its concepts." $\endgroup$ – Bill Dubuque Oct 18 '12 at 19:29
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    $\begingroup$ @user42912 The answer linked in my prior comment does - in passing - address some of your concerns. I think you will find it quite interesting (perhaps even entertaining). $\endgroup$ – Bill Dubuque Oct 18 '12 at 19:35

To make this completely formal without having to worry about what the variable $x$ means, we can define the ring of polynomials over some given ring $R$ to be the following:

The underlying set consists of all sequences of elements from $R$ (starting with the index $0$ for convenience), which are $0$ (ie, the $0$-element in $R$) from some point.

The addition is given term-wise and the multiplication is given by the following:

If we write $ab = c$ then $$c_i = \sum_{k + l = i}a_kb_l$$

This of course gives us exactly the same rules for addition and multiplication that we are used to, and it does not really help with anything but formalism (for example, the only way I can remember how the multiplication goes is through the usual way of writing polynomials).

Some additions:

As mentioned, we can see $x$ as the sequence $(0,1,0,\dots)$.

If $\alpha\in R$ and $f = (f_0,f_1,\dots,f_n,0,\dots)$ is some element in the ring of polynomials, then we can define the element $f(\alpha)\in R$ by $f(\alpha) = \sum_{i=0}^nf_i\alpha^i$ which then corresponds to the usual way we would evaluate a polynomial in the element $\alpha$.

  • $\begingroup$ That's the beginning of an explanation, but no cigar from me yet. At the very least, you should explain what $x$ is (answer: $x=(0,1,0,0,\ldots)$), and why. It might also be fun to note that this is actually a convolution algebra. $\endgroup$ – Harald Hanche-Olsen Oct 18 '12 at 19:21
  • $\begingroup$ @HaraldHanche-Olsen yes, so what is $f(\alpha)$? I mean, we can see F[x] as the set of the sequences $(a_0, a_1, a_3, ...)$ where all except a finite number of elements are 0. OK, I understand that, but based in this definition what is $f(\alpha)$? $\endgroup$ – user42912 Oct 18 '12 at 19:23
  • $\begingroup$ $\displaystyle\sum_{n=0}^{\infty}\:\left(a_n \cdot \alpha^n\right) \;\;$, $\;\;$ i.e., the output of the "evaluate at $\alpha$" homomorphism from $F\hspace{.01 in}[x]$ to $F$. $\hspace{0.4 in}$ $\endgroup$ – user57159 Oct 18 '12 at 19:38
  • $\begingroup$ @user42912: you can infer how “x” is encoded from evaluation formula. Unlike Fredrik Meyer’s stuff that is a pseudo-answer, this answer really explains something, although it possibly has shortcomings. $\endgroup$ – Incnis Mrsi Dec 12 '14 at 10:57

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