Find an example of an infinite algebraic extension over the field of rational numbers $\Q$ other than the algebraic closure $\bar{\Q}$ of $\Q$ in $\C$.

Definition (Algebraic Element, Algebraic Extension).

Let $F$ be a field and let $E$ be an extension of $F$.

• The element $\alpha \in E$ is said to be algebraic over $F$ is $\alpha$ is a root of some nonzero polynomial with coefficients in $F$.
• The extension $E/F$ is said to be algebraic if every element of $E$ is algebraic over $F$.

Proof.

Consider the field
\[K=\Q(\sqrt[3]{2}, \sqrt[5]{2}, \dots, \sqrt[2n+1]{2}, \dots).\] That is, $K$ is the field extension obtained by adjoining all numbers of the form $\sqrt[2n+1]{2}$ for any positive integers $n$.

Note that $\sqrt[2n+1]{2}$ is a root of the monic polynomial $x^{2n+1}-2$, hence $\sqrt[2n+1]{2}$ is algebraic over $\Q$.

By Eisenstein’s criterion with prime $2$, we know that the polynomial $x^{2n+1}-2$ is irreducible over $\Q$.
Thus the extension degree is $[\Q(\sqrt[2n+1]{2}):\Q]=2n+1$.

Since the field $K$ contains the subfield $\Q(\sqrt[2n+1]{2})$, we have
\[2n+1=[\Q(\sqrt[2n+1]{2}):\Q] \leq [K:\Q]\] for any positive integer $n$.
Therefore, the extension degree of $K$ over $\Q$ is infinite.

Observe that any element $\alpha$ of $K$ belongs to a subfield $\Q(\sqrt[3]{2}, \sqrt[5]{2}, \dots, \sqrt[2n+1]{2})$ for some $n \in \Z$.
Since each number $\sqrt[2k+1]{2}$ is algebraic over $\Q$, we know that this subfield is algebraic, hence $\alpha$ is algebraic.
Thus, the field $K$ is algebraic over $\Q$.

Is $K$ different from $\bar{\Q}$?

It remains to show that $K\neq \bar{\Q}$.

Consider $\sqrt{2}$.
Since $\sqrt{2}$ is a root of $x^2-2$, it is algebraic, hence $\sqrt{2}\in \bar{\Q}$.

We claim that $\sqrt{2}\not \in K$.
Assume on the contrary that $\sqrt{2} \in K$.
Then $\sqrt{2} \in F:=\Q(\sqrt[3]{2}, \sqrt[5]{2}, \dots, \sqrt[2n+1]{2}) \subset K$ for some $n \in \Z$.

Note that the extension degree of this subfield $F$ is odd since each extension degree of $\Q(\sqrt[2k+1]{2})/\Q$ is odd.

Since $\sqrt{2}\in F$, we must have
\begin{align*}
[F:\Q]=[F:\Q(\sqrt{2})][\Q(\sqrt{2}):\Q]=2[F:\Q(\sqrt{2})],
\end{align*}
which is even.

This is a contradiction, and hence $\sqrt{2}\not \in K$.
Thus, $K\neq \bar{\Q}$.

Comment.

With the same argument, we can prove that the field
\[K=\Q(\sqrt[2]{2}, \sqrt[3]{2}, \dots, \sqrt[n]{2}, \dots)\] is infinite algebraic extension over $\Q$.

However, it is not trivial to show that this field is different from $\bar{\Q}$.

That’s why we used only $\sqrt[2k+1]{2}$ in $K$.

We can also use $\sqrt[p]{2}$ for odd prime $p$.

The post Example of an Infinite Algebraic Extension first appeared on Problems in Mathematics.

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.

Let $\alpha= \sqrt[3]{2}e^{2\pi i/3}$. Prove that $x_1^2+\cdots +x_k^2=-1$ has no solutions with all $x_i\in \Q(\alpha)$ and $k\geq 1$.

Proof.

Note that $\alpha= \sqrt[3]{2}e^{2\pi i/3}$ is a root of the polynomial $x^3-2$.
The polynomial $x^3-2$ is irreducible over $\Q$ by Eisenstein’s criterion with prime $p=2$.

The roots of this polynomial are
\[\sqrt[3]{2}, \sqrt[3]{2}e^{2\pi i/3}, \sqrt[3]{2}e^{4\pi i/3}.\] Then it follows that we have an isomorphism
\begin{align*}
\Q(\alpha)=\Q(\sqrt[3]{2}e^{2\pi i/3}) \cong \Q[x]/(x^3-2)\cong \Q(\sqrt[3]{2}).
\end{align*}
Let us denote this isomorphism by $\phi:\Q(\alpha) \to \Q(\sqrt[3]{2})$, which sends $\alpha$ to $\sqrt[3]{2}$ and fixed $\Q$ elementwise.

Seeking a contradiction, we assume that there exist $x_1, \dots, x_k \in \Q(\alpha)$ such that
\[x_1^2+\cdots +x_k^2=-1.\]

Then we apply the isomorphism $\phi$ and obtain
\begin{align*}
-1&=\phi(-1)=\phi(x_1^2+\cdots +x_k^2)\\
&=\phi(x_1)^2+\cdots +\phi(x_k)^2.
\end{align*}

However, this equality does not hold since $\phi(x_i) \in \Q(\sqrt[3]{2})$ and the field $\Q(\sqrt[3]{2})$ consists of real numbers.

Thus, we have reached a contradiction, hence there is no solution of $x_1^2+\cdots +x_k^2=-1$ in the field $\Q(\alpha)$.

The post Equation $x_1^2+\cdots +x_k^2=-1$ Doesn't Have a Solution in Number Field $\Q(\sqrt[3]{2}e^{2\pi i/3})$ 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.

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 number $3$ divides all non-leading coefficients of $f(x)$. Also the constant term $6$ of $f(x)$ is not divisible by $3^2$. Hence by Eisenstein’s criterion, the polynomial $f(x)$ is irreducible over $\Q$.

We divide the polynomial $f(x)$ by $x+1$ and obtain
\[x^3+9x+6=(x+1)(x^2-x+10)-4\] by long division.

Then it follows that in the field $\Q(\theta) \cong \Q[x]/(f(x))$ (note that $f(x)$ is the minimal polynomial of $\theta$), we have
\[0=(\theta+1)(\theta^2-\theta+10)-4,\] and hence this yields that we have the inverse
\[(1+\theta)^{-1}=\frac{1}{4}(\theta^2-\theta+10).\] Click here if solved 60

The post Irreducible Polynomial $x^3+9x+6$ and Inverse Element in Field Extension 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.

Let $p \in \Z$ be a prime number.

Then describe the elements of the Galois group of the polynomial $x^p-2$.

Solution.

The roots of the polynomial $x^p-2$ are
\[ \sqrt[p]{2}\zeta^k, k=0,1, \dots, p-1\] where $\sqrt[p]{2}$ is a real $p$-th root of $2$ and $\zeta$ is a primitive $p$-th root of unity.
(Explicitly, you may take $\zeta=e^{2\pi i/p}$.)

Thus $x^p-2$ is a separable polynomial over $\Q$. The Galois group of $x^p-2$ is the Galois group of the splitting field of $x^p-2$.
The splitting field of $x^p-2$ is $K:=\Q(\sqrt[p]{2}, \zeta)$ of extension degree $p(p-1)$.
(Check this.)

Let $G=\Gal(K/\Q)$ be the Galois group of $x^p-2$. The order of the Galois group $G$ is $p(p-1)$. Let $\sigma \in G$ be an automorphism.
Then $\sigma$ sends an element of $G$ to its conjugate (a root of the minimal polynomial of the element.)
The minimal polynomial of $\sqrt[p]{2}$ is $x^p-2$ since it is irreducible by Eisenstein’s criteria.
The minimal polynomial of $\zeta$ is the cyclotomic polynomial
\[\Phi(x)=\frac{x^p-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+x+1.\]

Therefore $\sigma$ maps
\begin{align*}
\sqrt[p]{2} &\mapsto \sqrt[p]{2}\zeta^a \\
\zeta & \mapsto \zeta^b
\end{align*}
for some $a=0, 1, \dots, p-1$ and $b=1, 2, \dots, p-1$.

Thus there are $p(p-1)$ possible maps for $\sigma$.
Since the order of $G$ is $p(p-1)$, these are exactly the elements of the Galois group $G$ of the polynomial $x^p-2$.

The post Galois Group of the Polynomial $x^p-2$. first appeared on Problems in Mathematics.