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
Prove that $\sqrt[m]{2}$ is an irrational number for any integer $m \geq 2$.
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 […]

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 […]

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 […]

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, […]

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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 […]

Application of Field Extension to Linear Combination
Consider the cubic polynomial $f(x)=x^3-x+1$ in $\Q[x]$.
Let $\alpha$ be any real root of $f(x)$.
Then prove that $\sqrt{2}$ can not be written as a linear combination of $1, \alpha, \alpha^2$ with coefficients in $\Q$.
Proof.
We first prove that the polynomial […]