What if we consider polynomials whose coefficients are either rational or $e$, that is, a polynomial in $\mathbb{Q} \cup \{e\}$ with $\pi$ as a root.

Can this happen? Does it matter if we change rational to integer in the definition?

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    $\begingroup$ There is no difference between rational and integral for the coefficients (multiply through by denominators) $\endgroup$ – Mark Bennet Sep 6 '11 at 19:04
  • $\begingroup$ It's probably an open problem. The more general problem of algebraic independence is open. See mathoverflow.net/questions/33817/… $\endgroup$ – lhf Sep 6 '11 at 19:04

A real/complex number is a root of a polynomial with rational coefficients if and only if it is a root of a polynomial with integer coefficients: just multiply by a suitable integer to clear denominators to get an integer polynomial.

(However, it is not true that a complex number is a root of a monic polynomial with integer coefficients if and only if it is a root of a monic polynomial with rational coefficients; the former are called algebraic integers, and form a proper subset of the algebraic numbers. For example, $\frac{1}{2}$ is certainly the root of a monic polynomial with rational coefficients, e.g., $x - \frac{1}{2}$, but it's not the root of any monic polynomial with integer coefficients.)

Now, of course $\pi$ is a root of many polynomials with real coefficients: it's a root of $x-\pi$, after all, and so on.

Whether $\pi$ is a root of a polynomial with coefficients in $\mathbb{Q}[e]$ or in $\mathbb{Q}(e)$, however, is unknown. The notion you want is that of algebraic independence. A set of real numbers $\{r_1,\ldots,r_n\}$ is said to be "algebraically independent" if there does not exist a nonzero polynomial $p$ in $\mathbb{Q}[x_1,\ldots,x_n]$ such $p(r_1,\ldots,r_n)=0$. It is unknown whether $\{\pi,e\}$ is algebraically independent. (It's not even known whether $\pi+e$ is irrational!)

However, it is known that $\pi$ and $e^{\pi}$ are algebraically independent.

  • $\begingroup$ Is $\mathbb{Q}[e]$ the same as $\mathbb{Q} \cup \{e\}$? $\endgroup$ – Pedro Sep 6 '11 at 21:17
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    $\begingroup$ @Pedro: No. $\mathbb{Q} [e]$ consist of all real numbers which can be expressed as a polynomial in $e$ with rational coefficients. $\endgroup$ – Chris Eagle Sep 6 '11 at 21:29
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    $\begingroup$ @Pedro: not exactly; it's the set of 'polynomials in $e$' with coefficients in $\mathbb{Q}$; things like $e$, $e^2$, $6e^3-({1\over 2})e+{2\over 5}$, etc. $\endgroup$ – Steven Stadnicki Sep 6 '11 at 21:31
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    $\begingroup$ @Pedro: As others have noted, no; $\mathbb{Q}[e]$ is larger than $\mathbb{Q}\cup\{e\}$ (just like $\mathbb{Z}[\frac{1}{2}]$, which is the set of all rational numbers whose denominator is a power a power $2$ is strictly larger than $\mathbb{Z}\cup\{\frac{1}{2}\}$. But if $\pi$ is a polynomial with coefficients in $\mathbb{Q}\cup\{e\}$, then it is certainly one in with coefficients in $\mathbb{Q}[e]$. $\endgroup$ – Arturo Magidin Sep 6 '11 at 23:50

The answer is almost certainly no, but no proof is known. In fact, it is not even known whether $\pi e$ is rational. Schanuel's conjecture implies that $\pi$ and $e$ are algebraically independent, and therefore that the answer to your question is no.

  • $\begingroup$ @jsepcter: You mean, a joke? Sorry, then. (We still say Faltings proved "the Mordell conjecture", even though it's not a conjecture any more...) $\endgroup$ – Arturo Magidin Sep 6 '11 at 19:39

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