Joy of Mathematics

The integral domain is a field

November 10, 2020 Tolaso

Prove that an integral domain with the property that every strictly decreasing chain of ideals must be finite in length is a field.

On the centralizer

March 24, 2020March 24, 2020 Tolaso

Suppose that $A \in \mathcal{M}_n(\mathbb{C})$ has this property that if $\lambda$ is an eigenvalue of $A$ then $-\lambda$ is not an eigenvalue of $A$ . Show that $AX=XA$ if and only if $A^2X=XA^2$ for any $X \in \mathcal{M}_n(\mathbb{C})$ . In other words the centralizer of $A$ equals the centralizer of $A^2$ .

Solution

It is clear that $AX=XA$ implies $A^2X=XA^2$ for any $X \in \mathcal{M}_n(\mathbb{C})$ . Now suppose that $A^2X=XA^2$ for some $X \in \mathcal{M}_n(\mathbb{C})$ and set $Y=AX-XA$ . We want to prove that $Y=0$ . We have

$\begin{align*}AY+YA &=A(AX-XA)+(AX-XA)A\\ &=A^2X-XA^2\\ &=0 \end{align*}$

and so $AY=-YA$ . It now follows that $A^kY=(-1)^kYA^k$ for any integer $k \geq 0$ and thus for any $\lambda \in \mathbb{C}$ and any integer $m \geq 0$ we have

$\begin{align*}(A+\lambda \mathbb{I})^mY &=\sum_{k=0}^m\binom{m}{k}\lambda^{m-k}A^kY\\ &=\sum_{k=0}^m\binom{m}{k}(-1)^k\lambda^{m-k}YA^k \\ &=(-1)^mY\sum_{k=0}^m\binom{m}{k}(-\lambda)^{m-k}A^k \\ &=(-1)^mY(A-\lambda \mathbb{I}_{n})^m \quad (*) \end{align*}$

where $\mathbb{I}$ is the identity matrix. Now let $v$ be a generalized eigenvector corresponding to an eigenvalue $\lambda$ of $A$ . Then $(A-\lambda \mathbb{I}_{n})^mv=0$ for some integer $m$ and thus, by $(*)$ we have $(A+\lambda \mathbb{I}_n)^mYv=0$ . Therefore, since we are assuming that $-\lambda$ is not an eigenvalue of $A$ , we must have $Yv=0$ . So, since every element of $\mathbb{C}^n$ is a linear combination of some generalized eigenvectors of $A$ , we get $Yu=0$ for all $u \in \mathbb{C}^n$ , i.e. $Y=0$ and hence $AX=XA$ .

The exercise can also be found here.

Isomorphic groups

August 18, 2019August 18, 2019 Tolaso

Let $n>2$ . Define the group

$\mathcal{Q}_{2^n} = \langle x, y \mid x^2=y^{2^{n-2}} , y^{2^{n-1}} = 1, x^{-1} yx =y^{-1} \rangle$

Prove that $\displaystyle \mathcal{Q}_{2^n} / \mathcal{Z} \left ( \mathcal{Q}_{2^n} \right ) \simeq \mathcal{D}_{2^{n-1}}$ where $\mathcal{D}$ is the dihedral group.

Solution

Using $x^{-1}yx = y^{-1}$ or equivalently $yx = xy^{-1}$ we can write each element of $\mathcal{Q}_{2^n}$ in the form $x^ry^s$ where $r,s \in \mathbb{N} \cup \{0\}$ . Using $x^2 = y^{2^{n-2}}$ we may assume that $r\in \{0,1\}$ . Using $y^{2^{n-1}} = 1$ we may also assume that $s\in \{0,1,\ldots,2^{n-1}-1\}$ . It is easy to prove inductively that $y^tx = xy^{-t}$ .

Let $\mathcal{Z} = \mathcal{Z}(\mathcal{Q}_{2^n})$ . We prove that $\mathcal{Z}= \{1,y^{2^{n-2}}\}$ . Obviously $1 \in \mathcal{Z}$ . Furthermore, $y^{2^{n-2}} \in \mathcal{Z}$ since

$y^{2^{n-2}}\left(x^ry^s\right) = xy^{-2^{n-2}}x^{r-1}y^s = \cdots = x^ry^{2^{n-2}}y^s =\left(x^ry^s\right)y^{2^{n-2}}$

If $y^k \in \mathcal{Z}$ (such that $0 \leq k < 2^{n-1}$ ) then $xy^k = y^kx = xy^{-k}$ hence $y^{2k} = 1$ and therefore $k = 0$ or $k = 2^{n-2}$ . If $xy^k \in \mathcal{Z}$ then $xy^{k+1} = yxy^{k} = xy^{k-1}$ hence $y^2 = 1$ which is a contradiction since $n \geq 2$ .

Therefore,

$\mathcal{Q}_{2^n}/\mathcal{Z} = \langle x,y| x^2 = y^{2^{n-2}}=1, yx = xy^{-1} \rangle$

which is precisely the dihedral group with $2^{n-1}$ elements.

Inequality on groups

August 12, 2019 Tolaso

Let $\mathcal{G}$ be a finite group and suppose that $\mathcal{H}$ , $\mathcal{K}$ are two subgroups of $\mathcal{G}$ such that $\mathcal{H} \neq \mathcal{G}$ and $\mathcal{K} \neq \mathcal{G}$ . Show that

$\left|\mathcal{H} \cup \mathcal{K} \right| \leq \frac{3}{4} \left| \mathcal{G} \right|$

Solution

Recall that $\displaystyle |\mathcal{H}\mathcal{K}|=\frac{|\mathcal{H}||\mathcal{K}|}{|\mathcal{H} \cap \mathcal{K}|}$ and thus $\displaystyle \frac{|\mathcal{H}||\mathcal{K}|}{|\mathcal{H} \cap \mathcal{K}|} \leq |\mathcal{G}|$ . Hence $\displaystyle |\mathcal{H} \cap \mathcal{K}| \ge \frac{|\mathcal{H}| |\mathcal{K}|}{|\mathcal{G}|}$ and so

(1) $\begin{align*} |\mathcal{H} \cup \mathcal{K}|&=|\mathcal{H}|+|\mathcal{K}|-|\mathcal{H} \cap \mathcal{K}| \\ &\leq |\mathcal{H}|+|\mathcal{K}|-\frac{|\mathcal{H}| |\mathcal{K}|}{|\mathcal{G}|} \\ & =(a+b-ab)|\mathcal{G}| \end{align*}$

where $\displaystyle a=\frac{|\mathcal{H}|}{|\mathcal{G}|}$ and $\displaystyle b=\frac{|\mathcal{\mathcal{K}}|}{|\mathcal{G}|}$ .

Now, since $\mathcal{H} \neq \mathcal{G}$ and $\mathcal{K} \neq \mathcal{G}$ we have $[\mathcal{G}: \mathcal{H}] \geq 2$ and $[\mathcal{G}:\mathcal{K}] \geq 2$ that is $a \leq \frac{1}{2}$ and $b \leq \frac{1}{2}$ . So if we let $a'=1-2a$ and $b'=1-2b$ then $a', b' \geq 0$ and thus

$a+b-ab=\frac{3}{4}-\frac{a'+b'+a'b'}{4} \leq \frac{3}{4}$

due to $(1)$ .

The exercise along its solution have been migrated from here .

Irreducible factors of a polynomial

August 3, 2019August 3, 2019 Tolaso

Let $n \geq 1$ and let

$p_n(x)=x^{2^n}+x^{2^{n-1}}+1 \in \mathbb{Z}[x]$

Find all irreducible factors of $p_n(x)$ .

Solution

Setting $q_n(x)=x^{2^n}-x^{2^{n-1}}+1$ we note that

$p_n(x)=p_{n-1}(x)q_{n-1}(x) \quad , \quad n \geq 2$

Hence

(1) $\begin{equation*} p_n(x)=p_1(x)q_1(x)q_2(x) \cdots q_{n-1}(x) \end{equation*}$

It’s clear that $p_1(x)=x^2+x+1$ is irreducible over $\mathbb{Z}$ . Now, for $n \geq 1$ let $\Phi_n(x)$ be the $n$ -th cyclotomic polynomial. Using well-known properties of $\Phi_n$ , we have

$\Phi_{3 \cdot 2^n}(x)=\frac{\Phi_{2^n}(x^3)}{\Phi_{2^n}(x)}=\frac{x^{3 \cdot 2^{n-1}}+1}{x^{2^{n-1}}+1}=x^{2^n}-x^{2^{n-1}}+1=q_n(x)$

Thus $q_n$ is irreducible over $\mathbb{Z}$ because cyclotomic polynomials are irreducible over $\mathbb{Z}$ . Hence, by ( 1 ) $p_n$ has exactly $n$ irreducible factors and they are $p_1, q_1, q_2, \dots , q_{n-1}$ .