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Is the set of polynomial with coefficients on $\mathbb{Q}$ enumerable?

The set of integer coefficient polynomials are countable, when the cardinality of each set of length n polynomial is intepreted as $\mathbb{Z}^n$ for some finite n, then union of countable sets is countable.

What about this method, what is wrong with this ?

Suppose $P(x)=a_{0} + a_1{x} + a_2 x^2+..........$ is an infinite length polynomial, for each coefficient $a_i$ we have $\mathbb{Z}$ possible choices, so for $\mathbb{Z}$ terms we may choose $|\mathbb{Z}|^\mathbb{Z}$ possible polynomials and since $|\mathbb{Z}|^\mathbb{Z} > 2^\mathbb{Z}$ the set of integer coefficient polynomials is uncountable.

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    @Thomas Andrews, Yes I meant the polynomial algebra not a single polynomial per se.2012-09-15

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An 'infinite length polynomial' is not a polynomial, it is a (formal) power series.

Edit: Also $\left| \mathbb{Z} \right|^{\left| \mathbb{Z} \right|} = 2^{\left| \mathbb{Z} \right|}$, not $>$.

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    In particular, we often say the fundamential theorem of algebra is that any non-constant polynomial has a complex root. The inifinite power series, $1+x+x^2+... = \frac{1}{1-x}$ has not roots.2012-09-14