I am having trouble seeing how to apply the definition of transcendental to see this. Thanks!
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5Suppose $\pi$ has a multiplicative inverse, and see what happens. – Chris Eagle Oct 31 '12 at 21:21
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4In general, for $\alpha \in \mathbb C$, we have $\mathbb Q[\alpha]$ is a field iff $\alpha$ is algebraic. – lhf Oct 31 '12 at 21:26
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If $\mathbb{Q}[\pi]$ were a field, then $\pi$ would have an inverse. Every element in $\mathbb{Q}[\pi]$ is of the form $r_0+r_1\pi+r_2\pi^2+\cdots+r_n\pi^n$. So an inverse of $\pi$ would cause $(r_0+r_1\pi+r_2\pi^2+\cdots+r_n\pi^n)\pi=1$. But this is not possible, as this would imply $\pi$ is root of a polynomial with rational coeefficients, which it is not (it is transcendental).
Fredrik Meyer
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Hint $\ $ Notice $\:\pi\:$ transcendental over $\rm\Bbb Q\:\Rightarrow\:\Bbb Q[\pi]\cong \Bbb Q[x].\:$ But a polynomial ring cannot be a field since if $\rm\ x^{-1}\! = f(x)\in\Bbb Q[x]\ $ then $\rm\ x \; f(x) = 1 \: \Rightarrow\: 0 = 1,\ $ by evaluating at $\rm\ x = 0. $
Remark $\ $ The above proof has a very instructive universal interpretation.
Bill Dubuque
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Very clean. Would an isomorphism between the two rings be given by the map $\varphi(\pi) = x$, and $ \varphi(q) = q$ for all other $q \in \mathbb Q[\pi]$? – Moderat Nov 01 '12 at 02:58
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@jmi4 The kernel of the evaluation hom $\rm\: x\to \pi\:$ from $\rm\:\Bbb Q[x]\:$ onto $\rm\:\Bbb Q[\pi]\:$ is the ideal of polynomials $\rm\:f\in\Bbb Q[x]\:$ with $\rm\:f(\pi) = 0;\:$ if it contained any $\rm\:f \ne 0\:$ then $\,\pi\,$ would be algebraic over $\,\Bbb Q.\,$ Thus $\rm\:f = 0,\,$ i.e. the evaluation map has kernel $= 0,\:$ so it is an isomorphism. Informally: any transcendental element serves as an indeterminate. – Bill Dubuque Nov 01 '12 at 03:51