Give an example of a commutative ring $R$ and a prime ideal $I$ of $R$ that is not a maximal ideal of $R$.

Solution.

We give several examples. The key facts are:

1. An ideal $I$ of $R$ is prime if and only if $R/I$ is an integral domain.
2. An ideal $I$ of $R$ is maximal if and only if $R/I$ is a field.

Example 1: $\Z$ and $(0)$

The first example is the ring of integers $R=\Z$ and the zero ideal $I=(0)$.
Note that the quotient ring is $\Z/(0)\cong \Z$ and it is integral domain but not a field.
Thus the ideal $(0)$ is a prime ideal by Fact 1 but not a maximal ideal by Fact 2.

Remark

Note that $(0)$ is the only prime ideal of $\Z$ that is not a maximal ideal.
Nonzero ideals of $\Z$ are $(p)$ for some prime number $p$.

Example 2: $\Z[x]$ and $(x)$

The second example is the ring of polynomials $R=\Z[x]$ over $\Z$ and the principal ideal $I=(x)$ generated by $x\in \Z[x]$.
The quotient ring is $\Z[x]/(x)\cong \Z$, which is an integral domain but not a field.
Thus the ideal $(x)$ is prime but not maximal by Fact 1, 2.

Example 3: $\Q[x,y]$ and $(x)$

The third example is the ring of polynomials in two variables $R=\Q[x, y]$ over $\Q$ and the principal ideal $I=(x)$ generated by $x$.
The quotient ring $\Q[x,y]/(x)$ is isomorphic to $\Q[y]$.
(The proof of this isomorphism is given in the post Prove the Ring Isomorphism $R[x,y]/(x) \cong R[y]$.)

Note that $\Q[y]$ is an integral domain but it is not a field since, for instance, the element $y\in \Q[y]$ is not a unit.
Hence Fact 1, 2 implies that the ideal $(x)$ is prime but not maximal in the ring $\Q[x, y]$.

The post Examples of Prime Ideals in Commutative Rings that are Not Maximal Ideals first appeared on Problems in Mathematics.

Give an example of two groups $G$ and $H$ and a subgroup $K$ of the direct product $G\times H$ such that $K$ cannot be written as $K=G_1\times H_1$, where $G_1$ and $H_1$ are subgroups of $G$ and $H$, respectively.

Solution.

Let $G$ be any nontrivial group, and let $G=H$.
(For example, you may take $G=H=\Zmod{2}$.)

Then consider the subset $K$ in the direct product given by
\[K:=\{(g,g) \mid g\in G\} \subset G\times G.\]

We claim that $K$ is a subgroup of $G\times G$.
In fact, we have
\begin{align*}
(g,g)(h,h)=(gh,gh)\in K \text{ and }\\
(g,g)^{-1}=(g^{-1}, g^{-1})\in K
\end{align*}
for any $g, h\in G$.
Thus, $K$ is closed under multiplications and inverses, and hence $K$ is a subgroup of $G\times G$.

Now we show that $K$ is not of the form $G_1\times H_1$ for some subgroups $G_1, H_1$ of $G$.
Assume on the contrary $K=G_1\times H_1$ for some subgroups $G_1, H_1$ of $G$.

Since $G$ is a nontrivial group, there is a nonidentity element $x\in G$.
So $(x,x)\in K$ and $K$ is not the trivial group.
Thus, both $G_1$ and $H_1$ cannot be the trivial group.

Without loss of generality, assume that $G_1$ is nontrivial.
Then $G_1$ contains a nonidentity element $y$.

Since the identity element $e$ is contained in all subgroups, we have
\[(y,e)\in G_1\times H_1.\] However, this element cannot be in $K$ since $y\neq e$, a contradiction.

Hence $K$ is not of the form $G_1\times H_1$.

The post Example of Two Groups and a Subgroup of the Direct Product that is Not of the Form of Direct Product first appeared on Problems in Mathematics.

(a) Prove that each complex $n\times n$ matrix $A$ can be written as
\[A=B+iC,\] where $B$ and $C$ are Hermitian matrices.

(b) Write the complex matrix
\[A=\begin{bmatrix}
i & 6\\
2-i& 1+i
\end{bmatrix}\] as a sum $A=B+iC$, where $B$ and $C$ are Hermitian matrices.

Definition (Hermitian matrix).

Recall that a complex matrix $M$ is said to be Hermitian if $M^*=M$.
Here $A^*$ is the conjugate transpose matrix $M^*=\bar{M}^*$.

Proof.

Let
\[B=\frac{A+A^*}{2} \text{ and } C=\frac{A-A^*}{2i}.\] We claim that $B$ and $C$ are Hermitian matrices.
Using the fact that $(A^*)^*=A$, we compute
\begin{align*}
B^*&=\left(\, \frac{A+A^*}{2} \,\right)^*\\
&=\frac{A^*+(A^*)^*}{2}\\
&=\frac{A^*+A}{2}=B.
\end{align*}
It yields that the matrix $B$ is Hermitian.

We also have
\begin{align*}
C^*&=\left(\, \frac{A-A^*}{2i} \,\right)^*\\
&=\frac{A^*-(A^*)^*}{-2i}\\
&=\frac{A^*-A}{-2i}\\
&=\frac{A-A^*}{2i}=C.
\end{align*}
Thus, the matrix $C$ is also Hermitian.

Finally, note that we have
\begin{align*}
B+iC&=\frac{A+A^*}{2}+i\frac{A-A^*}{2i}\\
&=\frac{A+A^*}{2}+\frac{A-A^*}{2}\\
&=A.
\end{align*}
Therefore, each complex matrix $A$ can be written as $A=B+iC$, where $B$ and $C$ are Hermitian matrices.

\item By the proof of part (a), it suffices to compute
\[B=\frac{A+A^*}{2} \text{ and } C=\frac{A-A^*}{2i}.\]

We have
\[A^*=\begin{bmatrix}
-i & 2+i\\
6& 1-i
\end{bmatrix}.\]

A direct computation yields that
\[B=\begin{bmatrix}
0 & 4+\frac{i}{2}\\[6pt] 4-\frac{i}{2}& 1
\end{bmatrix} \text{ and } C=\begin{bmatrix}
1 & -\frac{1}{2}-2i\\[6pt] -\frac{1}{2}+2i& 1
\end{bmatrix}.\]

By the result of part (a), these matrices are Hermitian and satisfy $A=B+iC$, as required.

Related Question.

See the post “Express a Hermitian matrix as a sum of real symmetric matrix and a real skew-symmetric matrix” for a proof.

The post Every Complex Matrix Can Be Written as $A=B+iC$, where $B, C$ are Hermitian Matrices first appeared on Problems in Mathematics.

Find a nonzero $3\times 3$ matrix $A$ such that $A^2\neq O$ and $A^3=O$, where $O$ is the $3\times 3$ zero matrix.

(Such a matrix is an example of a nilpotent matrix. See the comment after the solution.)

Solution.

For example, let $A$ be the following $3\times 3$ matrix.
\[A=\begin{bmatrix}
0 & 1 & 0 \\
0 &0 &1 \\
0 & 0 & 0
\end{bmatrix}.\] Then $A$ is a nonzero matrix and we have
\[A^2=\begin{bmatrix}
0 & 1 & 0 \\
0 &0 &1 \\
0 & 0 & 0
\end{bmatrix}\begin{bmatrix}
0 & 1 & 0 \\
0 &0 &1 \\
0 & 0 & 0
\end{bmatrix}
=\begin{bmatrix}
0 & 0 & 1 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix}\neq O.\]

The third power of $A$ is
\[A^3=A^2A=\begin{bmatrix}
0 & 0 & 1 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix}\begin{bmatrix}
0 & 1 & 0 \\
0 &0 &1 \\
0 & 0 & 0
\end{bmatrix}=
\begin{bmatrix}
0 & 0 & 0 \\
0 &0 &0 \\
0 & 0 & 0
\end{bmatrix}=O.\] Thus, the nonzero matrix $A$ satisfies the required conditions $A^2\neq O, A^3=O$.

Comment.

A square matrix $A$ is called nilpotent if there is a non-negative integer $k$ such that $A^k$ is the zero matrix.
The smallest such an integer $k$ is called degree or index of $A$.

The matrix $A$ in the solution above gives an example of a $3\times 3$ nilpotent matrix of degree $3$.

The post Example of a Nilpotent Matrix $A$ such that $A^2\neq O$ but $A^3=O$. first appeared on Problems in Mathematics.