Prove $\mathbb{Q}(\zeta)$ is a field when $\zeta$ is algebraic

Question

Let $\zeta$ be a root of the irreducible rational polynomial $P(x)$ with degree $n$. I need to prove that the set of numbers of the form $S = \{ R(\zeta) | R(x) \in \mathbb{Q}[x] \}$ is a field.

I'm not sure of my solution.

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It's easily seen by piecewise addition/multiplication that $S$ is a ring, so to prove that $S$ is a field, we need to prove that inverses exist.

Pick any polynomial $R(x) \in \mathbb{Q}[x]$ - we will show the existence of $T(x) \in \mathbb{Q}[x]$ such that $T(\zeta)R(\zeta) = 1$.

For a polynomial $T(x) \in \mathbb{Q}[x]$, let $T'(x)$ be the polynomial with degree not more than $n-1$ such that $T'(x) \equiv T(x) \mod P(x)$.

Since $R(\zeta) = R'(\zeta)$, we can WLOG assume that $\deg[R] < n$.

By writing $T'(x) = \displaystyle \sum_{1 \leq i \leq n} a_ix^i$, consider the injection $f: \mathbb{Q}[x] \mapsto \mathbb{Q}^n$ such that $f(T(x)) = (a_0, a_1, \cdots, a_n)$.

For a polynomial $S(x)$ with degree not exceeding $n-1$, define the function $g_S: \mathbb{Q}^n \mapsto \mathbb{Q^n}$ such that $\displaystyle g_S(a_0, \cdots, a_n) = f((\sum_{1 \leq i \leq n} a_ix^i) S(x))$

Claim $g_S$ is a linear map with dimension equal to $n$
Proof: It's easy to see that $g_S$ is a linear mapping. Now we show that $\dim(g_S) = n$.

It suffices to show that kernel contains exactly one element, viz $0$. Assume $g_S(T) = (0, 0, \cdots, 0)$. Then $T(x)S(x) \equiv 0 \mod P(x) \Rightarrow R(x) | T(x)S(x)$. Now since $P(x)$ is irreducible, either $P(x) | S(x)$ or $R(x) | T(x)$. Since $ \deg[S] < \deg[R]$, the first case isn't possible, and the second isn't possible either if $T$ is nonzero. But it's true, so $T$ is zero polynomial, and we're done. $\blacksquare$

Now it follows that $g_s$ is invertible. Simply pick $T(x) = g^{-1}_S(1, 0, \cdots, 0)$ and we're done !

Answer 1

First of all you have to assume that $R(x)$ isn't divisible by $P(x)$, as otherwise we have that $R(\zeta) = 0$ in $\mathbb{Q}(\zeta)$ so it can't be invertible to begin with.Otherwise the proof seems alright at first look.

Nevertheless the easiest way would be to use the Extended Euclidean Algorithm. If we have that $P(x) \nmid R(x)$ then $\gcd(P(x),R(x)) = 1$ by the irreducibility of $P(x)$. Then by the Extended Euclidean Algorithm there exist $T,S \in \mathbb{Q}[x]$ s.t.

$$T(x)R(x) + S(x)P(x) = 1$$

As $P(\zeta) = 0$ in $\mathbb{Q}(\zeta)$ we get that $T(\zeta)R(\zeta) = 1 \in \mathbb{Q}(\zeta)$ and we are done.

Another way is to use the fact that $(P(x))$ generates a maximal ideal in $\mathbb{Q}[x]$ and using that $\mathbb{Q}(\zeta) \cong \mathbb{Q}[x]/(P(x))$ we get that $\mathbb{Q}(\zeta)$ is a field.

Answer 2

There is another very short way, using linear algebra:

Let's consider a non-zero element $p(\zeta)\in\mathbf Q[\zeta]$. Multiplication by $p(\zeta)$ in this $\mathbf Q$-vector space is injective since it is an integral domain (contained in $\mathbf C$). Now $\mathbf Q[\zeta]$ is a finite dimensional vector (with dimension $\deg P$), and an injective endomorphism of a finite dimensional vector space is surjective. In particular $1$ is attained, i.e. there exists polynomial $q$ such that $p(\zeta)q(\zeta)=1$ – in other words, $p(\zeta)$ is invertible in $\mathbf Q[\zeta]$.