Let $\F_3=\Zmod{3}$ be the finite field of order $3$.
Consider the ring $\F_3[x]$ of polynomial over $\F_3$ and its ideal $I=(x^2+1)$ generated by $x^2+1\in \F_3[x]$.

(a) Prove that the quotient ring $\F_3[x]/(x^2+1)$ is a field. How many elements does the field have?

(b) Let $ax+b+I$ be a nonzero element of the field $\F_3[x]/(x^2+1)$, where $a, b \in \F_3$. Find the inverse of $ax+b+I$.

(c) Recall that the multiplicative group of nonzero elements of a field is a cyclic group.

Confirm that the element $x$ is not a generator of $E^{\times}$, where $E=\F_3[x]/(x^2+1)$ but $x+1$ is a generator.

Proof.

(a) Prove that the quotient ring $\F_3[x]/(x^2+1)$ is a field

Let $f(x)=x^2+1$. We claim that the polynomial $f(x)$ is irreducible over $\F_3$.
To see this, note that $f(x)$ is a quadratic polynomial.
So $f(x)$ is irreducible over $\F_3$ if it does not have a root in $\F_3$.
We have
\begin{align*}
f(0)=1, \quad f(1)=2, \quad f(2)=2^2+1=2 \text{ in } \F_3.
\end{align*}
Hence $f(x)$ does not have a root in $\F_3$ and it is irreducible over $\F_3$.

It follows that the quotient $\F_3[x]/(x^2+1)$ is a field.
Since $x^2+1$ is quadratic, the extension degree of $\F_3[x]/(x^2+1)$ over $\F_3$ is $2$.
Hence the number of elements in the field is $3^2=9$.

(b) Find the inverse of $ax+b+I$

Let $ax+b$ be a representative of a nonzero element of the field $\F_3[x]/(x^2+1)$.
Let $cx+d$ be its inverse. Then we have
\begin{align*}
1&=(ax+b)(cx+d)=acx^2+(ad+bc)x+bd\\
&=(ad+bc)x+bd-ac
\end{align*}
since $x^2=-1$ in $\F_3[x]/(x^2+1)$.

Hence we obtain two equations
\begin{align*}
ad+bc=0 \text{ and } bd-ac=1.
\end{align*}

Since $ax+b$ is a nonzero element, at least one of $a, b$ is not zero.
If $a\neq 0$, then the first equation gives
\[d=-\frac{bc}{a}. \tag{*}\] Substituting this to the second equation, we obtain
\begin{align*}
\left(\, \frac{-b^2-a^2}{a} \,\right)c=1.
\end{align*}
Observe that $a^2+b^2$ is not zero in $\F_3$.
(Since $a \neq 0$, we have $a^2=1$. Also $b^2=0, 1$.)
Hence we have
\begin{align*}
c=-\frac{a}{a^2+b^2}.
\end{align*}

It follows from (*) that
\[d=\frac{b}{a^2+b^2}\]

Thus, if $a \neq 0$, then the inverse element is
\[(ax+b)^{-1}=\frac{1}{a^2+b^2}(-ax+b). \tag{**}\]

If $a=0$, then $b\neq 0$ and it is clear that the inverse element of $ax+b=b$ is $1/b$.
Note that the formula (**) is still true in this case.

In summary, we have

for any nonzero element $ax+b$ in the field $\F_3[x]/(x^2+1)$.

(c) $x$ is not a generator but $x+1$ is a generator

Note that the order of $E^{\times}$ is $8$ since $E$ is a finite field of order $9$ by part (a).
We compute the powers of $x$ and obtain
\begin{align*}
x, \quad x^2=-1, \quad x^3=-x, \quad x^4=-x^2=1.
\end{align*}
Thus, the order of the element $x$ is $4$, hence $x$ is not a generator of the cyclic group $E^{\times}$.

Next, let us check that $x+1$ is a generator.
We compute the powers of $x+1$ as follows.
\begin{align*}
&x+1, \quad (x+1)^2=x^2+2x+1=2x, \\
&(x+1)^3=2x(x+1)=2x^2+2x=2x-2=2x+1\\
&(x+1)^4=(2x+1)(x+1)=2x^2+3x+1=2.
\end{align*}

Observe that at this post the order of $x+1$ must be larger than $4$.
Since the order of $E^{\times}$ is $8$, the order of $x+1$ must be $8$ by Lagrange’s theorem.

Just for a reference we give the complete list of powers of $x+1$.
\[\begin{array}{ |c|c|}
\hline
n & (x+1)^n \\
\hline
1 & x+1 \\
2 & 2x \\
3 & 2x+1 \\
4 & 2 \\
5 & 2x+2\\
6 & x\\
7 &x+2\\
8 & 1\\
\hline
\end{array}\] Click here if solved 59

The post Prove that $\F_3[x]/(x^2+1)$ is a Field and Find the Inverse Elements first appeared on Problems in Mathematics.

Let $F$ be a finite field.
Prove that each element in the field $F$ is the sum of two squares in $F$.

Proof.

Let $x$ be an element in $F$. We want to show that there exists $a, b\in F$ such that
\[x=a^2+b^2.\]

Since $F$ is a finite field, the characteristic $p$ of the field $F$ is a prime number.

If $p=2$, then the map $\phi:F\to F$ defined by $\phi(a)=a^2$ is a field homomorphism, hence it is an endomorphism since $F$ is finite.( The map $\phi$ is called the Frobenius endomorphism).

Thus, for any element $x\in F$, there exists $a\in F$ such that $\phi(a)=x$.
Hence $x$ can be written as the sum of two squares $x=a^2+0^2$.

Now consider the case $p > 2$.
We consider the map $\phi:F^{\times}\to F^{\times}$ defined by $\phi(a)=a^2$. The image of $\phi$ is the subset of $F$ that can be written as $a^2$ for some $a\in F$.

If $\phi(a)=\phi(b)$, then we have
\[0=a^2-b^2=(a-b)(a+b).\] Hence we have $a=b$ or $a=-b$.
Since $b \neq 0$ and $p > 2$, we know that $b\neq -b$.
Thus the map $\phi$ is a two-to-one map.

Thus, there are $\frac{|F^{\times}|}{2}=\frac{|F|-1}{2}$ square elements in $F^{\times}$.
Since $0$ is also a square in $F$, there are
\[\frac{|F|-1}{2}+1=\frac{|F|+1}{2}\] square elements in the field $F$.

Put
\[A:=\{a^2 \mid a\in F\}.\] We just observed that $|A|=\frac{|F|+1}{2}$.

Fix an element $x\in F$ and consider the subset
\[B:=\{x-b^2 \mid b\in F\}.\] Clearly $|B|=|A|=\frac{|F|+1}{2}$.

Observe that both $A$ and $B$ are subsets in $F$ and
\[|A|+|B|=|F|+1 > |F|,\] and hence $A$ and $B$ cannot be disjoint.

Therefore, there exists $a, b \in F$ such that $a^2=x-b^2$, or equivalently,
\[x=a^2+b^2.\]

Hence each element $x\in F$ is the sum of two squares.

The post Each Element in a Finite Field is the Sum of Two Squares first appeared on Problems in Mathematics.

Prove that any field automorphism of the field of real numbers $\R$ must be the identity automorphism.

Proof.

We prove the problem by proving the following sequence of claims.

Let $\phi:\R \to \R$ be an automorphism of the field of real numbers $\R$.

1. Claim 1. For any positive real number $x$, we have $\phi(x)>0$.
2. Claim 2. For any $x, y\in \R$ such that $x>y$, we have $\phi(x) > \phi(y)$.
3. Claim 3. The automorphism $\phi$ is the identity on positive integers.
4. Claim 4. The automorphism $\phi$ is the identity on rational numbers.
5. Claim 5. The automorphism $\phi$ is the identity on real numbers.

Let us now start proving the claims.
Let $\phi:\R \to \R$ be an automorphism of the field of real numbers $\R$.

Claim 1. For any positive real number $x$, we have $\phi(x)>0$.

Since $x$ is a positive real number, we have $\sqrt{x}\in \R$ and
\[\phi(x)=\phi\left(\sqrt{x}^2\right)=\phi(\sqrt{x})^2 \geq 0.\]

Note that since $\phi(0)=0$ and $\phi$ is bijective, $\phi(x)\neq 0$ for any $x\neq 0$.
Thus, it follows that $\phi(x) > 0$ for each positive real number $x$.
Claim 1 is proved.

Claim 2. For any $x, y\in \R$ such that $x>y$, we have $\phi(x) > \phi(y)$.

Since $x > y$, we have $x-y > 0$ and it follows from Claim 1 that
\[0<\phi(x-y)=\phi(x)-\phi(y).\] Hence, $\phi(x)> \phi(y)$.

Claim 3. The automorphism $\phi$ is the identity on positive integers.

Let $n$ be a positive integer. Then we have
\begin{align*}
\phi(n)=\phi(\underbrace{1+1+\cdots+1}_{\text{$n$ times}})=\underbrace{\phi(1)+\phi(1)+\cdots+\phi(1)}_{\text{$n$ times}}=n
\end{align*}
since $\phi(1)=1$.

Claim 4. The automorphism $\phi$ is the identity on rational numbers.

Any rational number $q$ can be written as $q=\pm m/n$, where $m, n$ are positive integers.
Then we have
\begin{align*}
\phi(q)=\phi\left(\, \pm \frac{m}{n} \,\right)=\pm \frac{\phi(m)}{\phi(n)}=\pm \frac{m}{n}=q,
\end{align*}
where the third equality follows from Claim 3.

Claim 5. The automorphism $\phi$ is the identity on real numbers.

In this claim, we finish the proof of the problem.

Let $x$ be any real number.
Seeking a contradiction, assume that $\phi(x)\neq x$.

There are two cases to consider:
\[x < \phi(x) \text{ or } x > \phi(x).\]

First, suppose that $x < \phi(x)$. Then there exists a rational number $q$ such that \[x< q < \phi(x).\] Then we have \begin{align*} \phi(x) &< \phi(q) && \text{by Claim 2 since $x < q$}\\ &=q && \text{by Claim 4 since $q$ is rational}\\ &<\phi(x) && \text{by the choice of $q$}, \end{align*} and this is a contradiction. Next, consider the case when $x > \phi(x)$.
There exists a rational number $q$ such that
\[\phi(x) < q < x.\] Then by the same argument as above, we have \[\phi(x) < q =\phi(q) < \phi(x),\] which is a contradiction. Thus, in either case we reached a contradiction, and hence we must have $\phi(x)=x$ for all real numbers $x$. This proves that the automorphism $\phi: \R \to \R$ is the identity map.

The post Any Automorphism of the Field of Real Numbers Must be the Identity Map first appeared on Problems in Mathematics.

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.

Let $\zeta_8$ be a primitive $8$-th root of unity.
Prove that the cyclotomic field $\Q(\zeta_8)$ of the $8$-th root of unity is the field $\Q(i, \sqrt{2})$.

Proof.

Recall that the extension degree of the cyclotomic field of $n$-th roots of unity is given by $\phi(n)$, the Euler totient function.
Thus we have
\[[\Q(\zeta_8):\Q]=\phi(8)=4.\]

Without loss of generality, we may assume that
\[\zeta_8=e^{2 \pi i/8}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i.\]

Then $i=\zeta_8^2 \in \Q(\zeta_8)$ and $\zeta_8+\zeta_8^7=\sqrt{2}\in \Q(\zeta_8)$.
Thus, we have
\[\Q(i, \sqrt{2}) \subset \Q(\zeta_8).\]

It suffices now to prove that $[\Q(i, \sqrt{2}):\Q]=4$.
Note that we have $[\Q(i):\Q]=[\Q(\sqrt{2}):\Q]=2$.
Since $\Q(\sqrt{2}) \subset \R$, we know that $i\not \in \Q(\sqrt{2})$.
Thus, we have
\begin{align*}
[\Q(i, \sqrt{2}):\Q]=[[\Q(\sqrt{2})(i):\Q(\sqrt{2})][\Q(\sqrt{2}):\Q]=2\cdot 2=4.
\end{align*}

It follows that
\[\Q(\zeta_8)=\Q(i, \sqrt{2}).\] Click here if solved 24

The post The Cyclotomic Field of 8-th Roots of Unity is $\Q(\zeta_8)=\Q(i, \sqrt{2})$ first appeared on Problems in Mathematics.

Suppose that $\alpha$ is a rational root of a monic polynomial $f(x)$ in $\Z[x]$.
Prove that $\alpha$ is an integer.

Proof.

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.

The post A Rational Root of a Monic Polynomial with Integer Coefficients is an Integer 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.

Prove that any algebraic closed field is infinite.

Definition.

A field $F$ is said to be algebraically closed if each non-constant polynomial in $F[x]$ has a root in $F$.

Proof.

Let $F$ be a finite field and consider the polynomial
\[f(x)=1+\prod_{a\in F}(x-a).\] The coefficients of $f(x)$ lie in the field $F$, and thus $f(x)\in F[x]$. Of course, $f(x)$ is a non-constant polynomial.

Note that for each $a \in F$, we have
\[f(a)=1\neq 0.\] So the polynomial $f(x)$ has no root in $F$.
Hence the finite field $F$ is not algebraic closed.

It follows that every algebraically closed field must be infinite.

The post Prove that any Algebraic Closed Field is Infinite first appeared on Problems in Mathematics.

Let $n$ be an integer greater than $2$ and let $\zeta=e^{2\pi i/n}$ be a primitive $n$-th root of unity. Determine the degree of the extension of $\Q(\zeta)$ over $\Q(\zeta+\zeta^{-1})$.

The subfield $\Q(\zeta+\zeta^{-1})$ is called maximal real subfield.

Proof.

Note that since $n>2$, the primitive $n$-th root $\zeta$ is not a real number.
Also, we have
\begin{align*}
\zeta+\zeta^{-1}=2\cos(2\pi /n),
\end{align*}
which is a real number.

Thus the field $\Q(\zeta+\zeta^{-1})$ is real.
Therefore the degree of the extension satisfies
\[ [\Q(\zeta):\Q(\zeta+\zeta^{-1})] \geq 2.\]

We actually prove that the degree is $2$.
To see this, consider the polynomial
\[f(x)=x^2-(\zeta+\zeta^{-1})x+1\] in $\Q(\zeta+\zeta^{-1})[x]$.

The polynomial factos as
\[f(x)=x^2-(\zeta+\zeta^{-1})x+1=(x-\zeta)(x-\zeta^{-1}).\] Hence $\zeta$ is a root of this polynomial.

It follows from $[\Q(\zeta):\Q(\zeta+\zeta^{-1})] \geq 2$ that $f(x)$ is the minimal polynomial of $\zeta$ over $\Q(\zeta+\zeta^{-1})$, and hence the extension degree is
\[ [\Q(\zeta):\Q(\zeta+\zeta^{-1})] =2.\]

Comment.

The subfield $\Q(\zeta+\zeta^{-1})$ is called the maximal real subfield.
The reason why it is called as such should be clear from the proof.

The post Extension Degree of Maximal Real Subfield of Cyclotomic Field 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.