Proving That $\Bbb Q[\alpha]$ is a Field for all Algebraic Numbers $\alpha$ I'd like to show that $\alpha$ adjoined to $\mathbb{Q}$ is always a field, but I'm having trouble doing so.
For closure under addition, for example,
$(a_{0}+a_{1}\alpha +a_{2}\alpha ^{2}+...+a_{n-1}\alpha ^{n-1})+(b_{0}+b_{1}\alpha +b_{2}\alpha ^{2}+...+b_{n-1}\alpha ^{n-1})$
is clearly able to be expressed in the desired form.
Closure under multiplication isn't so simple, on the other hand. I'm not sure how to show that the product
$(a_{0}+a_{1}\alpha +a_{2}\alpha ^{2}+...+a_{n-1}\alpha ^{n-1})\times(b_{0}+b_{1}\alpha +b_{2}\alpha ^{2}+...+b_{n-1}\alpha ^{n-1})$
can also be expressed in the desired form and thus in $\mathbb{Q}[\alpha]$.
As for the multiplicative inverse, I'm also not sure how to prove that $\frac{1}{(a_{0}+a_{1}\alpha +a_{2}\alpha ^{2}+...+a_{n-1}\alpha ^{n-1})}$ is always able to be rationalized, and it seems like a pain to do so.
At this point, we have only been made aware of the definition of a field, so I suspect that a proof of this would follow a line of reasoning not necessarily based on just the definition of a field. The one hint given says to consider polynomials, but that doesn't help me out much. Help and hints would be appreciated.
 A: The set $\mathbb{Q}[\alpha]$ is the image of the ring homomorphism $\varphi\colon\mathbb{Q}[x]\to\mathbb{C}$ defined by
$$
\varphi(b_0+b_1x+\dots+b_nx^n)=b_0+b_1\alpha+\dots+b_n\alpha^n
$$
and therefore it is a subring.
Verifying that this is a homomorphism is exactly what you're attempting to do in order to show $\mathbb{Q}[\alpha]$ is a subring. However, this homomorphism is very useful for the rest of the proof.
More generally, if $\tau\colon R\to S$ is a homomorphism of commutative rings and $s\in S$ is a fixed element, there exists a unique ring homomorphism $\tau_s\colon R[x]\to S$ such that

*

*$\tau_s(r)=r$, for $r\in R$;

*$\tau_s(x)=s$.

The homomorphism is defined by $\tau_s(b_0+b_1x+\dots+b_nx^n)=\tau(b_0)+\tau(b_1)s+\dots+\tau(b_n)s^n$. In your case $\tau$ is the embedding $\mathbb{Q}\to\mathbb{C}$ and $s=\alpha$.
Now you surely know that $\alpha$ is the root of an irreducible polynomial $f(x)\in\mathbb{Q}[x]$ (the minimal polynomial). Let $c\in\mathbb{Q}[\alpha]$, $c\ne0$. Then
$$
c=\varphi(g(x))
$$
for some polynomial $g(x)\in\mathbb{Q}[\alpha]$ and, since $c\ne0$, we can deduce that $g(x)$ is not divisible by $f(x)$. Since $f(x)$ is irreducible, there are polynomials $p(x),q(x)\in\mathbb{Q}[\alpha]$ such that
$$
g(x)p(x)+f(x)q(x)=1
$$
because Bézout's identity carries over to $\mathbb{Q}[x]$, which is a Euclidean domain. Thus
$$
1=\varphi(1)=\varphi(g(x)p(x)+f(x)q(x))=g(\alpha)p(\alpha)+f(\alpha)q(\alpha)=g(\alpha)p(\alpha)=cp(\alpha)
$$
and so $c^{-1}=p(\alpha)\in\mathbb{Q}[\alpha]$.
