Existence of Multiplicative Inverse in a Commutative Ring

[Nirjhor]

Let $\alpha$ be an algebraic number. Prove that the Euclidean domain $\mathbb{Q}[\alpha]$ is in fact a field.

In other words, let $\alpha$ be a root of the polynomial $P$ such that $P$ is irreducible over $\mathbb Q$. Let $n=\deg P$. Define $\mathbb{A}=\left\{\displaystyle\sum_{j=0}^{n-1} q_j\alpha^j\mid q_j\in\mathbb{Q}~\forall j\right\}$. Prove that if $0\neq \gamma\in\mathbb{A}$ then $\gamma^{-1}\in\mathbb{A}$.

[Fm Jakaria]

In the polynomial ring $\mathbb{Q}[x]$ of field $\mathbb{Q}$, consider the principal ideal $I = <P(x)>$ generated by P[P is the minimal polynomial for $\alpha$]. Also consider the quotient ring $M = \mathbb{Q}[x]/I$; which is in fact a field since $P$ is irreducible.

Now note that the map $H : M \rightarrow \mathbb{Q}[\alpha]$ such that $A(x)+I \rightarrow A(\alpha)$ is well-defined, and is an isomorphism between the domain ring and codomain ring.
For the first, suppose $A(x)+I = B(x)+I$ with $A,B \in \mathbb{Q}[x]$.Then $A(x) – B(x) = C(x)P(x)…(i)$ for some $C \in$ $\mathbb{Q}[x]$. Now substitute $\alpha$ in (i) to see that $A(\alpha) = B(\alpha)$.
For the second, $H((A*B)(x)+I) = (A*B)(\alpha) = (A(\alpha))*(B(\alpha)) = H(A(x)+I) * H(B(x)+I)$; for any $A,B$ in $\mathbb{Q}[x]$, where $*$ is either ring addition or ring multiplication for both domain ring and codomain ring. Also $H(T(x)+I) = T(\alpha) = 1$, where $T$ is the constant polynomial $1$. So ring operations and multiplicative inverses are preserved, giving $H$ to be a homomorphism.
Now for any $x \in \mathbb{Q}[\alpha]$, for some polynomial $A \in \mathbb{Q}[x]; x = A(\alpha) = H(A(x)+I)$. So $H$ is surjective.
Also, if $A(\alpha) = B(\alpha)$ for some $A,B \in \mathbb{Q}[x]$, then $\alpha$ is a root of $C(x) = A(x) – B(x)$.
Write $C(x) = P(x)D(x) + R(x)$ for
$R = 0$ or $deg R < deg P$. The last is impossible, as $R$ has root $\alpha$, and $P$ is the minimal polynomial of $\alpha$. So $C(x) \in I$, so
$A(x) + I = B(x) + I$. So H is injective.
We proved that $H$ is an isomorphism. We conclude that as $M$ is a field, so is $\mathbb{Q}[\alpha]$.

[Nirjhor]

A Quicker Solution: Define the followings \[\sum_{j=0}^{n-1} q_j\alpha^j=\gamma\in\mathbb{A},~~P(x)=\sum_{j=0}^n a_jx^j,~~\Gamma(x)=\sum_{j=0}^{n-1}q_jx^j.\] Since $P$ is irreducible and $\deg P=1+\deg \Gamma$ it follows that $P$ and $\Gamma$ are coprime. Now by the extended Euclidean algorithm for polynomials, there exist polynomials $\Omega,\Psi\in\mathbb{Q}[x]$ so that \[P(x)\Omega(x)+\Gamma(x)\Psi(x)=\gcd\left(P,\Gamma\right)=1.\] Setting $x=\alpha$ and using $P(\alpha)=0$, $\Gamma(\alpha)=\gamma$ we have $\Psi(x)=\gamma^{-1}.$ Thus $\gamma^{-1}\in\mathbb{A}.$ Other closures hold trivially.