Prove that $S=\{f(\pi^5):f(x)\in \Bbb Q[x]\}$ is not a sub field of $\Bbb C$

Question

Put $S=\{f(\pi^5):f(x)\in \Bbb Q[x]\}$. Prove that $S$ is not a subfield of $\Bbb C$.

My attempt at a solution by contradiction, consider the 5th subfield axiom if $a$ has an inverse $b$ then $ab=1$ so now let $a=f(\pi^5)$ and $b=g(\pi^5)$ then $ab=f(\pi^5)g(\pi^5)=1$ but we know that $f(\pi^5)\ne 1$ as it is transcendent thus a contradiction

is this right? and if not can anyone guide me from here?

Answer 1

You're on the right track but you should argue as follows:

Let $a=\pi^5$. Then $a\in S$ and $a\ne0$. If there existed $b \in S$ such that $ab=1$, then we'd have $b=g(\pi^5)$ and $\pi^5g(\pi^5)=1$, which contradicts the fact that $\pi$ is transcendental because we'd have $h(\pi)=0$ for $h(x)=x^5g(x^5)-1 \in \mathbb Q[x]$, with $h \ne 0$.

Thus $a=\pi^5$ is an element of $S$ that does not have an inverse in $S$ and so $S$ is not a field.

Answer 2

Whatever be $f(x)\in \Bbb Q[x]$, you have $f(\pi^5)$ is a trascendental number because if not $\pi^5$ would be algebraic. Then all the elements of $S$ being trascendental, the number $1\notin S$ i.e. $S$ has not the identity for the group multiplicative hence can not be a field.

Answer 3

Let $a$ be a complex number. The map $$ e_a\colon\mathbb{Q}[x]\to \mathbb{C},\qquad e_a(f)=f(a) $$ is a ring homomorphism and its image is $S_a=\{f(a):f\in\mathbb{Q}[x]\}$, so $S_a$ is a subring of $\mathbb{C}$. Your case is $a=\pi^5$.

If $a$ is transcendental, then $\ker e_a=\{0\}$, so the homomorphism is injective and $S_a\cong\mathbb{Q}[x]$ as rings. Since $\mathbb{Q}[x]$ is not a field, neither $S_a$ is.

If $a$ is transcendental, then all of its powers $a^n$, with $n>0$ are transcendental. Indeed, if $g(a^n)=0$ for a nonzero polynomial $g\in\mathbb{Q}[x]$, we would have a nonzero polynomial $g_0\in\mathbb{Q}[x]$ with $g_0(a)=0$. So $\pi^5$ is transcendental, because $\pi$ is.