Suppose that $\alpha=\frac{p}{q}$ is a rational number in lowest terms, that is, $p$ and $q$ are relatively prime integers.

Let
\[f(x)=x^n+a_{n-1}x^{n-1}+\cdots+ a_1x+a_0\]
be a monic polynomial in $\Z[x]$ and $\alpha$ is a root of $f(x)$.

Since $\alpha$ is a root of $f(x)$, we have
\begin{align*}
0&=f(\alpha)\\
&=\alpha^n+a_{n-1}\alpha^{n-1}+\cdots+ a_1\alpha+a_0\\[6pt]
&=\frac{p^n}{q^n}+a_{n-1}\frac{p^{n-1}}{q^{n-1}}+\cdots+ a_1 \frac{p}{q}+a_0.
\end{align*}

Multiplying by $q^n$, we obtain
\begin{align*}
0&=q^n f(\alpha)\\
&=p^n+a_{n-1}qp^{n-1}+\cdots+ a_1 q^{n-1}p+a_0q^n\\[6pt]
&=p^n+q\left(\, a_{n-1}p^{n-1}+\cdots+ a_1 q^{n-2}p+a_0 q^{n-1} \,\right).
\end{align*}

Hence we have
\[q\left(\, a_{n-1}p^{n-1}+\cdots+ a_1 q^{n-2}p+a_0 q^{n-1} \,\right)=-p^n,\]
and this implies that $q$ divides $p^n$, and so $q$ divides $p$.

Since $p$ and $q$ are relatively primes, it yields that $q=1$.
Therefore, $\alpha=p/1=p$ is an integer.

$\sqrt[m]{2}$ is an Irrational Number
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Hint.
Use ring theory:
Consider the polynomial $f(x)=x^m-2$.
Apply Eisenstein's criterion, show that $f(x)$ is irreducible over $\Q$.
Proof.
Consider the monic polynomial […]

The Polynomial Rings $\Z[x]$ and $\Q[x]$ are Not Isomorphic
Prove that the rings $\Z[x]$ and $\Q[x]$ are not isomoprhic.
Proof.
We give three proofs.
The first two proofs use only the properties of ring homomorphism.
The third proof resort to the units of rings.
If you are familiar with units of $\Z[x]$, then the […]

Algebraic Number is an Eigenvalue of Matrix with Rational Entries
A complex number $z$ is called algebraic number (respectively, algebraic integer) if $z$ is a root of a monic polynomial with rational (respectively, integer) coefficients.
Prove that $z \in \C$ is an algebraic number (resp. algebraic integer) if and only if $z$ is an eigenvalue of […]

Show that Two Fields are Equal: $\Q(\sqrt{2}, \sqrt{3})= \Q(\sqrt{2}+\sqrt{3})$
Show that fields $\Q(\sqrt{2}+\sqrt{3})$ and $\Q(\sqrt{2}, \sqrt{3})$ are equal.
Proof.
It follows from $\sqrt{2}+\sqrt{3} \in \Q(\sqrt{2}, \sqrt{3})$ that we have $\Q(\sqrt{2}+\sqrt{3})\subset \Q(\sqrt{2}, \sqrt{3})$.
To show the reverse inclusion, […]

Degree of an Irreducible Factor of a Composition of Polynomials
Let $f(x)$ be an irreducible polynomial of degree $n$ over a field $F$. Let $g(x)$ be any polynomial in $F[x]$.
Show that the degree of each irreducible factor of the composite polynomial $f(g(x))$ is divisible by $n$.
Hint.
Use the following fact.
Let $h(x)$ is an […]

Polynomial $x^p-x+a$ is Irreducible and Separable Over a Finite Field
Let $p\in \Z$ be a prime number and let $\F_p$ be the field of $p$ elements.
For any nonzero element $a\in \F_p$, prove that the polynomial
\[f(x)=x^p-x+a\]
is irreducible and separable over $F_p$.
(Dummit and Foote "Abstract Algebra" Section 13.5 Exercise #5 on […]

Irreducible Polynomial $x^3+9x+6$ and Inverse Element in Field Extension
Prove that the polynomial
\[f(x)=x^3+9x+6\]
is irreducible over the field of rational numbers $\Q$.
Let $\theta$ be a root of $f(x)$.
Then find the inverse of $1+\theta$ in the field $\Q(\theta)$.
Proof.
Note that $f(x)$ is a monic polynomial and the prime […]