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Does any one how to prove that every entire algebraic function is a polynomial? I'm under the impression that this can be achieved by showing that an algebraic function grows no faster than a polynomial.

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    $\begingroup$ Not every entire function is a polynomial. For example, consider $e^z$. Could you be a bit more specific with your question? $\endgroup$ – Potato Feb 2 '12 at 0:32
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    $\begingroup$ The exponential function is entire and it is not a polynomial. $\endgroup$ – ncmathsadist Feb 2 '12 at 0:33
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    $\begingroup$ Do you mean every entire algebraic function is a polynomial? $\endgroup$ – Robert Israel Feb 2 '12 at 0:36
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    $\begingroup$ Based on the title, I believe the question is how to prove that every algebraic entire function is a polynomial. $\endgroup$ – Jonas Meyer Feb 2 '12 at 0:36
  • $\begingroup$ yes I'm sorry I meant every entire algebraic function is a polynomial. $\endgroup$ – Ricky Maher Feb 2 '12 at 0:40
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I assume you mean every entire algebraic function is a polynomial. Suppose $g$ is an entire algebraic function of order $n$. Thus there are polynomials $c_j(z)$, $j=0 \ldots, n$ with $c_n$ not identically $0$ such that $\sum_{j=0}^n c_j(z) g(z)^j = 0$. For all but finitely many complex numbers $w$, $\sum_{j=0}^n c_j(z) w^j$ is not identically $0$, and so there are at most finitely many $z$ for which $\sum_{j=0}^n c_j(z) w^j = 0$: those are the only $z$ for which we can have $g(z) = w$. Thus for all but finitely many $w$, $g(z)$ takes the value $w$ only finitely many times. But an entire function that is not a polynomial has an essential singularity at $\infty$, and by the Great Picard Theorem it takes all but one value infinitely many times.

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  • $\begingroup$ For what it is worth, I have a survey by V. I. Arnold that mentions the function $z(a,b,c)$ that satisfies $$z^7 + a z^3 + b z^2 + c z + 1 = 0$$ and says in a footnote that it is "an entire algebraic function (without poles)" and asks whether it can be expressed as a superposition of such functions in two variables. $\endgroup$ – Will Jagy Feb 2 '12 at 1:17
  • $\begingroup$ @Will, I noticed that one, too. I convinced myself that in the context of that piece, "entire algebraic function" doesn't mean "function that is both entire and algebraic," but something else. $\endgroup$ – Gerry Myerson Feb 2 '12 at 1:48
  • $\begingroup$ @GerryMyerson, thanks, I decided not to leave that as a comment to the OP partly because I was uncertain what was going on. Technical matter, it seems we can reply to a comment with merely an @ sign and a first name? Because the software did tell me about your comment, in the "responses" flag in my personal profile page. Hmmm, if I look at your profile, some things are not visible to me, such as your "response" flag, so I cannot quickly experiment with this. $\endgroup$ – Will Jagy Feb 2 '12 at 2:09
  • $\begingroup$ @Wil, my understanding is that the first three letters of the name suffice. $\endgroup$ – Gerry Myerson Feb 2 '12 at 4:44
  • $\begingroup$ @Ger, it worked. Thanks. $\endgroup$ – Will Jagy Feb 2 '12 at 5:05
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Hopefully not, because $\: \operatorname{exp} : \mathbb{C} \to \mathbb{C} \:$ defined by $\:\:\:\: f(z) \:\: = \:\: \displaystyle\sum_{n=0}^{\infty} \: \left(\frac1{n!} \cdot \left(z^n\right)\right)$
is an entire function that is not a polynomial.


See http://en.wikipedia.org/wiki/Exponential_function#Complex_plane.

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    $\begingroup$ See the title; I believe the functions are also supposed to be algebraic. $\endgroup$ – Jonas Meyer Feb 2 '12 at 0:37
  • $\begingroup$ so far i have the following: 1. If (1/abs value z)(abs value f) holds then f is a polynomial. So assume theorem not true... $\endgroup$ – Ricky Maher Feb 2 '12 at 0:43
  • $\begingroup$ @RickyMaher: "$\frac{1}{|z|}|f|$" is not a statement. What do you mean by "$\frac{1}{|z|}|f|$ holds"? $\endgroup$ – Jonas Meyer Feb 2 '12 at 0:51

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