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The following is from The Classical Groups Their Invariants and Representations, by Hermann Weyl:

$f\left[x\right]$ being a polynomial in $x,$ $\alpha$ is a root or zero of $f$ if $f\left[\alpha\right]=0.$ A polynomial of degree $n$ has at most $n$ different zeros; this follows in the well-known way by proving that $f\left[x\right]$ contains the factors $(x-\alpha_{1})(x-\alpha_{2})\dots$ if $\alpha_{1},\alpha_{2},\dots$ are distinct zeros. Hence a polynomial $f\left[x\right]\ne0$ does not vanish numerically for every value of $x$ in $k,$ provided the reference field $k$ is of characteristic $0,$ because such a field contains infinitely many numbers.

That statement seems problematic to me. It appears to say that a polynomial that is non-zero somewhere is not zero everywhere, which is a tautology. Is he really saying that a polynomial whose coefficients are not all zero will not vanish identically on a field of characteristic $0$?

That is either obviously the case, or I am failing to understand what is meant by a polynomial on a field of characteristic $0$.

Steven Thomas Hatton
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    It’s not quite so obvious when you realize that the polynomial $X^p-X$ is identically zero when considered as a function on the field $\Bbb F_p=\Bbb Z/p\Bbb Z$. – Lubin Jan 17 '19 at 03:09
  • I'm unfamiliar with your notation. Care to elaborate? – Steven Thomas Hatton Jan 17 '19 at 03:41
  • Apparently you are referring to a field with prime characteristic, which is not a field with characteristic 0. – Steven Thomas Hatton Jan 17 '19 at 04:06
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    You seem to be confused by the difference between a polynomial and a polynomial function. You see, a polynomial is simply a sum of finitely many terms of the form $a_ix^i$, where $a_i$ is a constant, $i$ is a non-negative integer, and $x$ is an indeterminate. See here or possibly here as well as other threads linked to those for some discussion. In high school and calculus applications we teach students to think of $x$ as taking values from some suitable set. – Jyrki Lahtonen Jan 17 '19 at 06:24
  • (cont'd) Choosing a value for $x$ from a set (=evaluating the polynomial) creates a polynomial function. While we think of $x$ as an indeterminate we have a formal polynomial. The theorems in abstract algebra (polynomial long division, Euclidean algorithm in the ring of polynomials, PID property, etc) really work with formal polynomials. Reread those passages (possibly with newly opened eyes) to see that! – Jyrki Lahtonen Jan 17 '19 at 06:30
  • @JyrkiLahtonen Well, the first book I checked, and the first sentence I laid eyes on calls Weyl's "polynomial expression" an "entire rational function mapping $R$ into itself". It goes onto state that if $f\ne{0}$ "we may obviously assume $a_n\ne{0}$." The domain is a commutative ring with unit element. That source goes on to give a different definition of "indeterminate" than what Weyl gives. They define indeterminate as synonymous with "transcendent". I will never be so bold as the claim that books on abstract algebra will be consistent in their definitions as usage. NEVER! – Steven Thomas Hatton Jan 17 '19 at 17:31
  • @JyrkiLahtonen Apologies. That is really more in response to Eric Wofsey than you. I am editing my original question to address the comments. – Steven Thomas Hatton Jan 17 '19 at 17:54

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Is he really saying that a polynomial whose coefficients are not all zero will not vanish identically on a field of characteristic 0?

Yes, that's exactly what he's saying. Note that a polynomial is defined as a sequence of coefficients, not as a function. So $f[x]\neq 0$ means that the sequence of coefficients is not the zero sequence, i.e. that the coefficients are not all zero.

Eric Wofsey
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  • That's not the definition Weyl gives. Nor is it a definition I have seen in use elsewhere. Weyl's definition is :A formal expression

    $f\left[x\right]=\sum_{i=0}^{n}\alpha_{i}x^{i}$

    involving the “indeterminate” (or variable) $x,$ whose coefficients $\alpha_{i}$ are numbers in a field $k$, is called a $\left(k-\right)\text{polynomial}$ of formal degree $n$.

     – Steven Thomas Hatton Jan 17 '19 at 10:45
  • The term "formal expression" means that it is defined by its sequence of coefficients, not as a function. This is completely standard and is the definition of a polynomial you will find in literally any abstract algebra text. – Eric Wofsey Jan 17 '19 at 16:27
  • In vector and tensor mathematics (which, ironically, I learned from Weyl), $\left{a_i\right}_0^n$ is the set of components of a vector on the basis $\left\langle x^{i}\right\rangle _{0}^{n}$. The statement $f\left[x\right]=0$ is a linear equation which implies $a_i=0,$ rather than defining it to be so. Wikipedia (which I don't take as authoritative) agrees with your interpretation https://en.wikipedia.org/wiki/Polynomial#Abstract_algebra So, I guess it is reasonable to accept it as Weyl's intent. – Steven Thomas Hatton Jan 17 '19 at 20:04