[Math] Why isn’t $\frac{1}{x}$ a polynomial

abstract-algebrapolynomialsring-theory

Why isn't $\frac{1}{x}$ a polynomial?

Does it directly follow from definition? As far as I know, polynomials in $F$ are expressions of the form $\sum_{i=0}^{n} a_ix^i$, where $a_i\in F$ and $x$ is a symbol.

Or is there a nicer argument involved?

Footnote: $F$ is a field of characteristic zero.

Best Answer

If you want a more formal "proof", you can suppose for contradiction that $1/x$ is in fact equal to some expression $a_k x^n$ of the form $ \sum_{k=0}^n a_k x^n $:

$$ \frac{1}{x} = a_0 + a_1 x + \cdots + a_n x^n$$

multiplying through by $x$ gives:

$$ 1 = a_0 x + a_1 x^2 + \cdots + a_n x^{n+1}$$

Setting $x=0$ gives: $$ 1 = a_0 \cdot 0 + \cdots + a_n \cdot 0^{n+1} = 0$$

a contradiction, so we must have that $1/x$ is not a polynomial.