# Are there irreducible polynomials of "infinite" degree?

If $\alpha$ is transcendental over $F$, we sometimes say that $[F(\alpha): F] =\infty$. I'm wondering if this is true in a literal sense, namely: is there some irreducible polynomial of "infinite" degree which has $\alpha$ as one of its roots?

My first attempt at formalizing this question was:

(*) Can we find some sequence of polynomials $f_1, f_2,\dots$ with roots $r_1, r_2,\dots$ such that $\lim_{n\to\infty} r_n =\alpha$?

Unfortunately, at least in some cases the answer to this question is "yes" for an uninteresting reason: if we have some Cauchy sequence $q_1, q_2,\dots$ which converges to $\alpha$, then $\{(x - q_i)\}_i$ meets (*) but isn't describing a polynomial of "infinite" degree in any meaningful sense.

So I'm wondering: is there some meaningful sense in which polynomials of infinite degree can have roots that polynomials of finite degree do not? And if so, can all transcendental numbers the written as the root of such polynomials?

• What do you mean by $\lim_{n \rightarrow \infty} r_n = \alpha$? Are you assuming there's some kind of topology? Sep 25, 2016 at 23:13
• You can't say that an infinite power series, evaluated at $\alpha$, is equal to $0$ unless you have some notion of convergence in your field. Thus basically the only field where this becomes useful is when $F$ and $F(\alpha)$ lie between $\Bbb Q$ and $\Bbb C$. You can probably shoehorn it into other fields, but I wouldn't recommend it until you're more experienced. Sep 25, 2016 at 23:14
• There is no such thing as a "polynomial of infinite degree", for much the same reason as there is no such thing as a triangle with four sides. There is such a thing as a power series, and if $\alpha$ is a transcendental number, then you can certainly find a power series $f$ with rational coefficients such that $f(\alpha)=0$. Try it! Sep 26, 2016 at 7:18