Why is $Q[\pi]$ not a field?

Solution 1:

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).

Solution 2:

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.