Prove that the cubic polynomial $x^3-2$ is irreducible over the field $\Q(i)$.

Proof.

Note that the polynomial $x^3-2$ is irreducible over $\Q$ by Eisenstein’s criterion (with prime $p=2$).
This implies that if $\alpha$ is any root of $x^3-2$, then the degree of the field extension $\Q(\alpha)$ over $\Q$ is $3$:
\[[\Q(\alpha) : \Q]=3. \tag{*}\]

Seeking a contradiction, assume that $x^3-2$ is reducible over $\Q(i)$.
Then $x^3-2$ has a root in $\Q(i)$ as it is a reducible degree $3$ polynomial. So let us call the root $\alpha \in \Q(i)$.

Then $\Q(\alpha)$ is a subfield of $\Q(i)$ and thus we have
\[2=[\Q(i) :\Q]=[\Q(i): \Q(\alpha)][\Q(\alpha):\Q]\geq 3\] by (*). Hence we have reached a contradiction.
As a result, $x^3-2$ is irreducible over $\Q(i)$.

The post Cubic Polynomial $x^3-2$ is Irreducible Over the Field $\Q(i)$ first appeared on Problems in Mathematics.

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 $f(x)=x^3-x+1$ is irreducible over $\Q$.
Since $f(x)$ is a monic cubic polynomial, the only possible roots are the divisors of the constant term $1$. As we have $f(1)=f(-1)=1\neq 0$, the polynomial has no rational roots. Hence $f(x)$ is irreducible over $\Q$.

Then $f(x)$ is the minimal polynomial of $\alpha$ over $\Q$, and hence the field extension $\Q(\alpha)$ over $\Q$ has degree $3$.
If $\sqrt{2}$ is a linear combination of $1, \alpha, \alpha^2$, then it follows that $\sqrt{2}\in \Q(\alpha)$. Then $\Q(\sqrt{2})$ is a subfield of $\Q(\alpha)$.

Then the degree of the field extension is
\begin{align*}
3=[\Q(\alpha): \Q]=[\Q(\alpha): \Q(\sqrt{2})] [\Q(\sqrt{2}): \Q].
\end{align*}
Since $[\Q(\sqrt{2}): \Q]=2$, this is impossible.
Thus, $\sqrt{2}$ is not a linear combination of $1, \alpha, \alpha^2$.

The post Application of Field Extension to Linear Combination first appeared on Problems in Mathematics.

Let $R$ be an integral domain and let $S=R[t]$ be the polynomial ring in $t$ over $R$. Let $n$ be a positive integer.

Prove that the polynomial
\[f(x)=x^n-t\] in the ring $S[x]$ is irreducible in $S[x]$.

Proof.

Consider the principal ideal $(t)$ generated by $t$ in $S$.
Then the ideal $(t)$ is a prime ideal in $S$ since the quotient
\[S/(t)=R[t]/(t)\cong R\] is an integral domain.

The only non-leading coefficient of $f(x)=x^n-t$ is $-t$, and $-t$ is in the ideal $(t)$ but not in the ideal $(t)^2$.
Then by Eisenstein’s criterion, the polynomial $f(x)$ is irreducible in $S[x]$.

(Remark that $S=R[t]$ is an integral domain since $R$ is an integral domain.)

The post Irreducible Polynomial Over the Ring of Polynomials Over Integral Domain first appeared on Problems in Mathematics.

Show that the polynomial
\[f(x)=x^4-2x-1\] is irreducible over the field of rational numbers $\Q$.

Proof.

We use the fact that $f(x)$ is irreducible over $\Q$ if and only if $f(x+a)$ is irreducible for any $a\in \Q$.
We prove that the polynomial $f(x+1)$ is irreducible.

We have
\begin{align*}
f(x+1)&=(x+1)^4-2(x+1)-1\\
&=(x^4+4x^3+6x^2+4x+1)-2(x+1)-1\\
&=x^4+4x^3+6x^2+2x-2.
\end{align*}
Then the polynomial $f(x+1)$ is monic and all the non-leading coefficients are divisible by the prime number $2$.

Since the constant term is not divisible by $2^2$, Eisenstein’s criterion implies that the polynomial $f(x+1)$ is irreducible over $\Q$.
Therefore by the fact stated above, the polynomial $f(x)$ is also irreducible over $\Q$.

The post Polynomial $x^4-2x-1$ is Irreducible Over the Field of Rational Numbers $\Q$ first appeared on Problems in Mathematics.

Show that $\Q(\sqrt{2+\sqrt{2}})$ is a cyclic quartic field, that is, it is a Galois extension of degree $4$ with cyclic Galois group.

Proof.

Put $\alpha=\sqrt{2+\sqrt{2}}$. Then we have $\alpha^2=2+\sqrt{2}$. Taking square of $\alpha^2-2=\sqrt{2}$, we obtain $\alpha^4-4\alpha^2+4=2$. Hence $\alpha$ is a root of the polynomial
\[f(x)=x^4-4x+2.\] By the Eisenstein’s criteria, $f(x)$ is an irreducible polynomial over $\Q$.

There are four roots of $f(x)$:
\[\pm \sqrt{2 \pm \sqrt{2}}.\] Note that we have a relation
\[(\sqrt{2+\sqrt{2}})(\sqrt{2-\sqrt{2}})=\sqrt{2}.\] Thus we have
\[\sqrt{2-\sqrt{2}}=\frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}} \in \Q(\sqrt{2+\sqrt{2}}).\]

Hence all the roots of $f(x)$ are in the field $\Q(\sqrt{2+\sqrt{2}})$, hence $\Q(\sqrt{2+\sqrt{2}})$ is the splitting field of the separable polynomial $f(x)=x^4-4x+2$.
Thus the field $\Q(\sqrt{2+\sqrt{2}})$ is Galois over $\Q$ of degree $4$.

Let $\sigma \in \Gal(\Q(\sqrt{2+\sqrt{2}})/ \Q)$ be the automorphism sending
\[\sqrt{2+\sqrt{2}} \mapsto \sqrt{2-\sqrt{2}}.\] Then we have
\begin{align*}
2+\sigma(\sqrt{2})&=\sigma(2+\sqrt{2})\\
&=\sigma\left((\sqrt{2+\sqrt{2}}) ^2 \right)\\
&=\sigma \left(\sqrt{2+\sqrt{2}} \right) ^2\\
&= \left(\sqrt{2-\sqrt{2}} \right)^2=2-\sqrt{2}.
\end{align*}
Thus we obtain $\sigma(\sqrt{2})=-\sqrt{2}$.

Using this, we have
\begin{align*}
\sigma^2(\sqrt{2+\sqrt{2}})&=\sigma(\sqrt{2-\sqrt{2}})\\
&=\sigma \left(\frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}} \right)\\
&=\frac{\sigma(\sqrt{2})}{\sigma(\sqrt{2+\sqrt{2}})} \\
&=\frac{-\sqrt{2}}{\sqrt{2-\sqrt{2}}} \\
&=-\sqrt{2-\sqrt{2}}.
\end{align*}
Therefore $\sigma^2$ is not the identity automorphism. This implies the Galois group $\Gal(\Q(\sqrt{2+\sqrt{2}})/ \Q)$ is generated by $\sigma$, that is, the Galois group is a cyclic group of order $4$.

The post Galois Extension $\Q(\sqrt{2+\sqrt{2}})$ of Degree 4 with Cyclic Group first appeared on Problems in Mathematics.

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

Hint.

Use ring theory:
1. Consider the polynomial $f(x)=x^m-2$.
2. Apply Eisenstein’s criterion, show that $f(x)$ is irreducible over $\Q$.

Proof.

Consider the monic polynomial $f(x)=x^m-2$ in $\Z[x]$.
The constant term is divisible by the prime $2$ and not divisible by $2^2$.

Thus, by Eisenstein’s criterion, the polynomial $f(x)$ is irreducible over the rational numbers $\Q$.
In particular, it does not have a degree $1$ factor.

If $\sqrt[m]{2}$ is rational, then $x-\sqrt[m]{2}\in Q[x]$ is a degree $1$ factor of $f(x)$ and this cannot happen.
Therefore, $\sqrt[m]{2}$ is an irrational number for any integer $m\geq 2$.

The post $\sqrt[m]{2}$ is an Irrational Number first appeared on Problems in Mathematics.

Prove that the polynomial $x^p-2$ for a prime number $p$ is irreducible over the field $\Q(\zeta_p)$, where $\zeta_p$ is a primitive $p$th root of unity.

Hint.

Consider the field extension $\Q(\sqrt[p]{2}, \zeta)$, where $\zeta$ is a primitive $p$-th root of unity.

Remark

The following proof proves more than enough. You might want to refine the proof for a simpler proof.

Proof.

We first determine the splitting field of $x^p-2$.
The roots of the polynomial $x^p-2$ are
\[\sqrt[p]{2}\zeta^i,\] where $\zeta$ is a primitive $p$-th root of unity and $i=0,1,\dots, p-1$.

Let $F$ be the splitting field of $x^p-2$. Since both $\sqrt[p]{2}$ and $\sqrt[p]{2}\zeta$ is in $F$, the quotient $\zeta=\sqrt[p]{2}\zeta/\sqrt[p]{2} \in F$.
Therefore we see that $\Q(\sqrt[p]{2}, \zeta)\subset F$. Since the field $\Q(\sqrt[p]{2}, \zeta)$ contains all the roots of $x^p-2$, we must have $F=\Q(\sqrt[p]{2}, \zeta)$.

Next, we find the degree of the extension $\Q(\sqrt[p]{2}, \zeta)$ over $\Q$.
The field $\Q(\sqrt[p]{2}, \zeta)$ contains the cyclotomic field $\Q(\zeta)$ as a subfield and we obtain $\Q(\sqrt[p]{2}, \zeta)$ by adjoining $\sqrt[p]{2}$ to $\Q(\zeta)$.
Since $\sqrt[p]{2}$ is a root of $x^p-2$, the degree of the extension $[\Q(\sqrt[p]{2}, \zeta) : \Q(\zeta)] \leq p$.

Thus we have
\[ [\Q(\sqrt[p]{2}, \zeta): \Q]=[\Q(\sqrt[p]{2}, \zeta): \Q(\zeta)] [\Q(\zeta): \Q]\leq p(p-1)\] since the degree of cyclotomic extension over $\Q$ is $\phi(p)=p-1$. (Here $\phi$ is the Euler phi function.)

Note that $\Q(\sqrt[p]{2})$ is also a subfield and $[\Q(\sqrt[p]{2}): \Q]=p$ since $x^p-2$ is irreducible over $\Q$ by Eisenstein’s criterion.
Hence both $p$ and $p-1$ divide $[\Q(\sqrt[p]{2}, \zeta): \Q] \leq p(p-1)$. Since $p$ is a prime, the numbers $p$ and $p-1$ are relatively prime. Thus we must have $[\Q(\sqrt[p]{2}, \zeta): \Q]=p(p-1)$.

In particular, the polynomial $x^p-2$ must be irreducible over $\Q(\zeta)$ otherwise the degree $[\Q(\sqrt[p]{2}, \zeta): \Q]$ is strictly less than $p(p-1)$.

The post The Polynomial $x^p-2$ is Irreducible Over the Cyclotomic Field of $p$-th Root of Unity first appeared on Problems in Mathematics.

Show that the polynomial $x^3-\sqrt{2}$ is irreducible over the field $\Q(\sqrt{2})$.

Hint.

Consider the field extensions $\Q(\sqrt{2})$ and $\Q(\sqrt[6]{2})$.

Proof.

Let $\sqrt[6]{2}$ denote the positive real $6$-th root of of $2$.
Then since $x^6-2$ is irreducible over $\Q$ by Eisenstein’s criterion, hence it is the minimal polynomial for $\sqrt[6]{2}$ over $\Q$. Therefore, we have $[\Q(\sqrt[6]{2}): \Q]=6$.
Similarly, we have $[\Q(\sqrt{2}):\Q]=2$.

Since $\Q(\sqrt[6]{2}) \ni (\sqrt[6]{2})^3=\sqrt{2}$ and $[\Q(\sqrt{2}):\Q]=2$, we must have
\[[\Q(\sqrt[6]{2}): \Q(\sqrt{2})]=3.\]

Note that $\sqrt[6]{2}$ is a root of the polynomial $x^3-\sqrt{2} \in \Q(\sqrt{2})[x]$.
Hence it must be the minimal polynomial for $\sqrt[6]{2}$ over $\Q(\sqrt{2})$.
In particular, the polynomial is irreducible over $\Q(\sqrt{2})$.

The post $x^3-\sqrt{2}$ is Irreducible Over the Field $\Q(\sqrt{2})$ first appeared on Problems in Mathematics.