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

Question

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.

Answer 1

Note that: $$\frac{1}{x}=x^{-1}$$

Going off of the wikipedia definition of a polynomial found here:

In mathematics, a polynomial is an expression consisting of variables (also called indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.

It is easy to see that our expression fails to meet the criteria of being a polynomial due to the fact that its variable contains a negative exponent.

Answer 2

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.

Answer 3

You can first use intuition from polynomials where coefficients are from the real numbers. Being every where continuous the image of $[-1,1]$ is a compact set, which is bounded for a polynomial. But the image of the same set under $f(x)= \frac1x$ is unbounded. So we can see that in reals, $1/x$ is not a polynomial.

Now take any field of characteristic 0. One can see that if $\alpha$ is any root of $g(x)$ then it has to be a root of the product polynomial $f(x)g(x)$. Now assuming $f(x)=\frac1x$ is a polynomial multiply with the second polynomial $g(x)=x(x-1)$. Their product is the polynomial $(x-1)$, this product has just 1 root namely $1$ whereas $g(x)$ has two roots, 0 and 1. This contradiction should settle the issue.

Answer 4

The short answer, as pointed out by Randall in his comment, is that polynomials are by definition sums of terms of the form $ax^k$ where $k \ge 0$; since $x^{-1}$ is not of this type, it is not polynomial. This actually covers the case of formally defined polynomials

$p(x) \in F[x], \tag 1$

since there is no term of the form $x^{-1} \in F[x]$ according to the conventional definition, which only addresses non-negative powers of $x$.

Perhaps a somewhat more subtle question is whether, as a function, $x^{-1}$ may be expressed an element of $F[x]$; that is, can we ever have

$x^{-1} = p(x) = \displaystyle \sum_0^n p_i x^i \in F[x], \; p_i \in F, \; 0 \le p_i \le n? \tag 2$

the usual understanding of this equation, as an equivalence of functions, is that

$\forall 0 \ne a \in F, \; a^{-1} = p(a). \tag 3$

Under the hypothesis that

$\text{char}(F) = 0 \tag 4$

we may rule (3) out as follows: it is equivalent to

$\forall 0 \ne a \in F, \; ap(a) = 1, \tag 5$

which in fact asserts that every $0 \ne a \in F$ is a root of the polynomial

$xp(x) - 1 = \displaystyle \sum_0^n p_i x^{i + 1} - 1; \tag 6$

we have

$\deg(xp(x) - 1) = n + 1; \tag 7$

as such, $xp(x) - 1$ has at most $n + 1$ zeroes in $F$; but (4) implies that $F$ contains a copy of the rationals $\Bbb Q$ as an infinite subfield; every non-zero element of $\Bbb Q$ must thus satisfy (6), and hence a reduction to absurdity is attained. Therefore, (2) cannot be the case. $OE\Delta.$

Answer 5

We can also show the result ignoring the usual construction of $F(X)$ and instead using (only) the universal property definition:

The polynomial ring $F[X]$ is a ring $P$, together with a ring homomorphism $i\colon F\to P$ and a special element $X\in P$, such that for all rings $A$, ring homomorphisms $f\colon F\to A$, and elements $a\in A$, there exists one and only one ring homomorphism $h\colon P\to A$ with $h\circ i=f$ and $h(X)=a$.

Now assume there exists $u\in P$ such that $uX=1$ (or, if we do not demand unital rings, just $uX=i(e)$ for some $e\ne0$). Consider $A=F$, $a=0$, $f=\operatorname{id}_F$. By the universal property, there exists $h\colon P\to F$ such that $h\circ i=\operatorname{id}_F$ and $h(X)=0$. Then $$e=h(i(e))=h(uX)=h(u)h(X)=h(u)0=0,$$ a contradiction.