# Is the set of integer coefficient polynomials countable? [duplicate]

Possible Duplicate:
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.

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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## marked as duplicate by Asaf Karagila, William, Thomas, J. M., NorbertOct 3 '12 at 20:08

There is a lovely bijection between $\mathbb{N}$ and $\mathbb{N}[X]$, namely, $\sum a_n x^n \rightarrow \prod p_n^{a_n}$ where $p_n$ is the nth prime. Of course, it is trivial to extend to $\mathbb{Z}$. As it was pointed out, polynomials are finite. – Karolis Juodelė Sep 14 '12 at 18:59
An integral polynomial gives a function from $\mathbb{Z} \rightarrow \mathbb{Z}$. A power series doesn't. A polynomial is a finitely generated algebra over the co-efficient ring a power series is not. – s.b Sep 14 '12 at 19:25
@s.b. Being pedantic, but "a polynomial" is not "a finitely generated algebra." The "set of polynomials" is "a finitely generated algebra." – Thomas Andrews Sep 14 '12 at 19:32
This is not an exact duplicate per se, but I think it is close enough. – Asaf Karagila Sep 15 '12 at 9:13
@AsafKaragila I would consider this an abstract duplicate since $|\mathbb{Z}|=|\mathbb{Q}|$ (but I can't vote...) – Belgi Sep 15 '12 at 16:18

Edit: Also $\left| \mathbb{Z} \right|^{\left| \mathbb{Z} \right|} = 2^{\left| \mathbb{Z} \right|}$, not $>$.
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. – Thomas Andrews Sep 14 '12 at 19:34