# A prime ideal that is not a maximal ideal

Every maximal ideal is a prime ideal. The converse is true in a principal ideal domain – PID, i.e. every nonzero prime ideal is maximal in a PID, but this is not true in general. Let’s produce a counterexample.

$$R= \mathbb Z[x]$$ is a ring. $$R$$ is not a PID as can be shown considering the ideal $$I$$ generated by the set $$\{2,x\}$$. $$I$$ cannot be generated by a single element $$p$$. If it was, $$p$$ would divide $$2$$, i.e. $$p=1$$ or $$p=2$$. We can’t have $$p=1$$ as it means $$R = I$$ but $$3 \notin I$$. We can’t have either $$p=2$$ as it implies the contradiction $$x \notin I$$. The ideal $$J = (x)$$ is a prime ideal as $$R/J \cong \mathbb Z$$ is an integral domain. Since $$\mathbb Z$$ is not a field, $$J$$ is not a maximal ideal.

# Four elements rings

A group with four elements is isomorphic to either the cyclic group $$\mathbb Z_4$$ or to the Klein four-group $$\mathbb Z_2 \times \mathbb Z_2$$. Those groups are commutative. Endowed with the usual additive and multiplicative operations, $$\mathbb Z_4$$ and $$\mathbb Z_2 \times \mathbb Z_2$$ are commutative rings.

Are all four elements rings also isomorphic to either $$\mathbb Z_4$$ or $$\mathbb Z_2 \times \mathbb Z_2$$? The answer is negative. Let’s provide two additional examples of commutative rings with four elements not isomorphic to $$\mathbb Z_4$$ or $$\mathbb Z_2 \times \mathbb Z_2$$.

The first one is the field $$\mathbb F_4$$. $$\mathbb F_4$$ is a commutative ring with four elements. It is not isomorphic to $$\mathbb Z_4$$ or $$\mathbb Z_2 \times \mathbb Z_2$$ as both of those rings have zero divisor. Indeed we have $$2 \cdot 2 = 0$$ in $$\mathbb Z_4$$ and $$(1,0) \cdot (0,1)=(0,0)$$ in $$\mathbb Z_2 \times \mathbb Z_2$$.

A second one is the ring $$R$$ of the matrices $$\begin{pmatrix} x & 0\\ y & x\end{pmatrix}$$ where $$x,y \in \mathbb Z_2$$. One can easily verify that $$R$$ is a commutative subring of the ring $$M_2(\mathbb Z_2)$$. It is not isomorphic to $$\mathbb Z_4$$ as its characteristic is $$2$$. This is not isomorphic to $$\mathbb Z_2 \times \mathbb Z_2$$ either as $$\begin{pmatrix} 0 & 0\\ 1 & 0\end{pmatrix}$$ is a non-zero matrix solution of the equation $$X^2=0$$. $$(0,0)$$ is the only solution of that equation in $$\mathbb Z_2 \times \mathbb Z_2$$.

One can prove that the four rings mentioned above are the only commutative rings with four elements up to isomorphism.

# Group homomorphism versus ring homomorphism

A ring homomorphism is a function between two rings which respects the structure. Let’s provide examples of functions between rings which respect the addition or the multiplication but not both.

### An additive group homomorphism that is not a ring homomorphism

We consider the ring $$\mathbb R[x]$$ of real polynomials and the derivation $\begin{array}{l|rcl} D : & \mathbb R[x] & \longrightarrow & \mathbb R[x] \\ & P & \longmapsto & P^\prime \end{array}$ $$D$$ is an additive homomorphism as for all $$P,Q \in \mathbb R[x]$$ we have $$D(P+Q) = D(P) + D(Q)$$. However, $$D$$ does not respect the multiplication as $D(x^2) = 2x \neq 1 = D(x) \cdot D(x).$ More generally, $$D$$ satisfies the Leibniz rule $D(P \cdot Q) = P \cdot D(Q) + Q \cdot D(P).$

### A multiplication group homomorphism that is not a ring homomorphism

The function $\begin{array}{l|rcl} f : & \mathbb R & \longrightarrow & \mathbb R \\ & x & \longmapsto & x^2 \end{array}$ is a multiplicative group homomorphism of the group $$(\mathbb R, \cdot)$$. However $$f$$ does not respect the addition.

# A group G isomorph to the product group G x G

Let’s provide an example of a nontrivial group $$G$$ such that $$G \cong G \times G$$. For a finite group $$G$$ of order $$\vert G \vert =n > 1$$, the order of $$G \times G$$ is equal to $$n^2$$. Hence we have to look at infinite groups in order to get the example we’re seeking for.

We take for $$G$$ the infinite direct product $G = \prod_{n \in \mathbb N} \mathbb Z_2 = \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 \dots,$ where $$\mathbb Z_2$$ is endowed with the addition. Now let’s consider the map $\begin{array}{l|rcl} \phi : & G & \longrightarrow & G \times G \\ & (g_1,g_2,g_3, \dots) & \longmapsto & ((g_1,g_3, \dots ),(g_2, g_4, \dots)) \end{array}$

From the definition of the addition in $$G$$ it follows that $$\phi$$ is a group homomorphism. $$\phi$$ is onto as for any element $$\overline{g}=((g_1, g_2, g_3, \dots),(g_1^\prime, g_2^\prime, g_3^\prime, \dots))$$ in $$G \times G$$, $$g = (g_1, g_1^\prime, g_2, g_2^\prime, \dots)$$ is an inverse image of $$\overline{g}$$ under $$\phi$$. Also the identity element $$e=(\overline{0},\overline{0}, \dots)$$ of $$G$$ is the only element of the kernel of $$G$$. Hence $$\phi$$ is also one-to-one. Finally $$\phi$$ is a group isomorphism between $$G$$ and $$G \times G$$.

# Isomorphism of factors does not imply isomorphism of quotient groups

Let $$G$$ be a group and $$H, K$$ two isomorphic subgroups. We provide an example where the quotient groups $$G / H$$ and $$G / K$$ are not isomorphic.

Let $$G = \mathbb{Z}_4 \times \mathbb{Z}_2$$, with $$H = \langle (\overline{2}, \overline{0}) \rangle$$ and $$K = \langle (\overline{0}, \overline{1}) \rangle$$. We have $H \cong K \cong \mathbb{Z}_2.$ The left cosets of $$H$$ in $$G$$ are $G / H=\{(\overline{0}, \overline{0}) + H, (\overline{1}, \overline{0}) + H, (\overline{0}, \overline{1}) + H, (\overline{1}, \overline{1}) + H\},$ a group having $$4$$ elements and for all elements $$x \in G/H$$, one can verify that $$2x = H$$. Hence $$G / H \cong \mathbb{Z}_2 \times \mathbb{Z}_2$$. The left cosets of $$K$$ in $$G$$ are $G / K=\{(\overline{0}, \overline{0}) + K, (\overline{1}, \overline{0}) + K, (\overline{2}, \overline{0}) + K, (\overline{3}, \overline{0}) + K\},$ which is a cyclic group of order $$4$$ isomorphic to $$\mathbb{Z}_4$$. We finally get the desired conclusion $G / H \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \ncong \mathbb{Z}_4 \cong G / K.$

# A Commutative Ring with Infinitely Many Units

In a ring $$R$$ a unit is any element $$u$$ that has a multiplicative inverse $$v$$, i.e. an element $$v$$ such that $uv=vu=1,$ where $$1$$ is the multiplicative identity.

The only units of the commutative ring $$\mathbb Z$$ are $$-1$$ and $$1$$. For a field $$\mathbb F$$ the units of the ring $$\mathrm M_n(\mathbb F)$$ of the square matrices of dimension $$n \times n$$ is the general linear group $$\mathrm{GL}_n(\mathbb F)$$ of the invertible matrices. The group $$\mathrm{GL}_n(\mathbb F)$$ is infinite if $$\mathbb F$$ is infinite, but the ring $$\mathrm M_n(\mathbb F)$$ is not commutative for $$n \ge 2$$.

The commutative ring $$\mathbb Z[\sqrt{2}] = \{a + b\sqrt{2} \ ; \ (a,b) \in \mathbb Z^2\}$$ is not a field. However it has infinitely many units.

### $$a + b\sqrt{2}$$ is a unit if and only if $$a^2-2b^2 = \pm 1$$

For $$u = a + b\sqrt{2} \in \mathbb Z[\sqrt{2}]$$ we denote $$\mathrm N(u) = a^2- 2b^2 \in \mathbb Z$$. For any $$u,v \in \mathbb Z[\sqrt{2}]$$ we have $$\mathrm N(uv) = \mathrm N(u) \mathrm N(v)$$. Therefore for a unit $$u \in \mathbb Z[\sqrt{2}]$$ with $$v$$ as multiplicative inverse, we have $$\mathrm N(u) \mathrm N(v) = 1$$ and $$\mathrm N(u) =a^2-2b^2 \in \{-1,1\}$$.

### The elements $$(1+\sqrt{2})^n$$ for $$n \in \mathbb N$$ are unit elements

The proof is simple as for $$n \in \mathbb N$$ $(1+\sqrt{2})^n (-1 + \sqrt{2})^n = \left((1+\sqrt{2})(-1 + \sqrt{2})\right)^n=1$

One can prove (by induction on $$b$$) that the elements $$(1+\sqrt{2})^n$$ are the only units $$u \in \mathbb Z[\sqrt{2}]$$ for $$u \gt 1$$.

# A normal extension of a normal extension may not be normal

An algebraic field extension $$K \subset L$$ is said to be normal if every irreducible polynomial, either has no root in $$L$$ or splits into linear factors in $$L$$.

One can prove that if $$L$$ is a normal extension of $$K$$ and if $$E$$ is an intermediate extension (i.e., $$K \subset E \subset L$$), then $$L$$ is a normal extension of $$E$$.

However a normal extension of a normal extension may not be normal and the extensions $$\mathbb Q \subset \mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt[4]{2})$$ provide a counterexample. Let’s prove it.

As a short lemma, we prove that a quadratic extension $$k \subset K$$ , i.e. an extension of degree two is normal. Suppose that $$P$$ is an irreducible polynomial of $$k[x]$$ with a root $$a \in K$$. If $$a \in k$$ then the degree of $$P$$ is equal to $$1$$ and we’re done. Otherwise $$(1, a)$$ is a basis of $$K$$ over $$k$$ and there exist $$\lambda, \mu \in k$$ such that $$a^2 = \lambda a +\mu$$. As $$a \notin k$$, $$Q(x)= x^2 – \lambda x -\mu$$ is the minimal polynomial of $$a$$ over $$k$$. As $$P$$ is supposed to be irreducible, we get $$Q = P$$. And we can conclude as $Q(x) = (x-a)(x- \lambda +a).$

The entensions $$\mathbb Q \subset \mathbb Q(\sqrt{2})$$ and $$\mathbb Q(\sqrt{2}) \subset \mathbb Q(\sqrt[4]{2})$$ are quadratic, hence normal according to previous lemma and $$\sqrt[4]{2}$$ is a root of the polynomial $$P(x)= x^4-2$$ of $$\mathbb Q[x]$$. According to Eisenstein’s criterion $$P$$ is irreducible over $$\mathbb Q$$. However $$\mathbb Q(\sqrt[4]{2}) \subset \mathbb R$$ while the roots of $$P$$ are $$\pm \sqrt[4]{2}, \pm i \sqrt[4]{2}$$ and therefore not all real. We can conclude that $$\mathbb Q \subset \mathbb Q(\sqrt[4]{2})$$ is not normal.

# The image of an ideal may not be an ideal

If $$\phi : A \to B$$ is a ring homomorphism then the image of a subring $$S \subset A$$ is a subring $$\phi(A) \subset B$$. Is the image of an ideal under a ring homomorphism also an ideal? The answer is negative. Let’s provide a simple counterexample.

Let’s take $$A=\mathbb Z$$ the ring of the integers and for $$B$$ the ring of the polynomials with integer coefficients $$\mathbb Z[x]$$. The inclusion $$\phi : \mathbb Z \to \mathbb Z[x]$$ is a ring homorphism. The subset $$2 \mathbb Z \subset \mathbb Z$$ of even integers is an ideal. However $$2 \mathbb Z$$ is not an ideal of $$\mathbb Z[x]$$ as for example $$2x \notin 2\mathbb Z$$.

# A group that is not a semi-direct product

Given a group $$G$$ with identity element $$e$$, a subgroup $$H$$, and a normal subgroup $$N \trianglelefteq G$$; then we say that $$G$$ is the semi-direct product of $$N$$ and $$H$$ (written $$G=N \rtimes H$$) if $$G$$ is the product of subgroups, $$G = NH$$ where the subgroups have trivial intersection $$N \cap H= \{e\}$$.

Semi-direct products of groups provide examples of non abelian groups. For example the dihedral group $$D_{2n}$$ with $$2n$$ elements is isomorphic to a semidirect product of the cyclic groups $$\mathbb Z_n$$ and $$\mathbb Z_2$$. $$D_{2n}$$ is the group of isometries preserving a regular polygon $$X$$ with $$n$$ edges.

Let’see that the converse is not true and present a group that is not a semi-direct product.

### The Hamilton’s quaternions group is not a semi-direct product

The Hamilton’s quaternions group $$\mathbb H_8$$ is the group consisting of the symbols $$\pm 1, \pm i, \pm j, \pm k$$ where$-1 = i^2 =j^2 = k^2 \text{ and } ij = k = -ji,jk = i = -kj, ki = j = -ik.$ One can prove that $$\mathbb H_8$$ endowed with the product operation above is indeed a group having $$8$$ elements where $$1$$ is the identity element.

$$\mathbb H_8$$ is not abelian as $$ij = k \neq -k = ji$$.

Let’s prove that $$\mathbb H_8$$ is not the semi-direct product of two subgroups. If that was the case, there would exist a normal subgroup $$N$$ and a subgroup $$H$$ such that $$G=N \rtimes H$$.

• If $$\vert N \vert = 4$$ then $$H = \{1,h\}$$ where $$h$$ is an element of order $$2$$ in $$\mathbb H_8$$. Therefore $$h=-1$$ which is the only element of order $$2$$. But $$-1 \in N$$ as $$-1$$ is the square of all elements in $$\mathbb H_8 \setminus \{\pm 1\}$$. We get the contradiction $$N \cap H \neq \{1\}$$.
• If $$\vert N \vert = 2$$ then $$\vert H \vert = 4$$ and $$H$$ is also normal in $$G$$. Noting $$N=\{1,n\}$$ we have for $$h \in H$$ $$h^{-1}nh=n$$ and therefore $$nh=hn$$. This proves that the product $$G=NH$$ is direct. Also $$N$$ is abelian as a cyclic group of order $$2$$. $$H$$ is also cyclic as all groups of order $$p^2$$ with $$p$$ prime are abelian. Finally $$G$$ would be abelian, again a contradiction.

We can conclude that $$G$$ is not a semi-direct product.

# A normal subgroup that is not a characteristic

Let’s $$G$$ be a group. A characteristic subgroup is a subgroup $$H \subseteq G$$ that is mapped to itself by every automorphism of $$G$$.

An inner automorphism is an automorphism $$\varphi \in \mathrm{Aut}(G)$$ defined by a formula $$\varphi : x \mapsto a^{-1}xa$$ where $$a$$ is an element of $$G$$. An automorphism of a group which is not inner is called an outer automorphism. And a subgroup $$H \subseteq G$$ that is mapped to itself by every inner automorphism of $$G$$ is called a normal subgroup.

Obviously a characteristic subgroup is a normal subgroup. The converse is not true as we’ll see below.

### Example of a direct product

Let $$K$$ be a nontrivial group. Then consider the group $$G = K \times K$$. The subgroups $$K_1=\{e\} \times K$$ and $$K_2=K \times \{e\}$$ are both normal in $$G$$ as for $$(e, k) \in K_1$$ and $$(a,b) \in G$$ we have
$(a,b)^{-1} (e,x) (a,b) = (a^{-1},b^{-1}) (e,x) (a,b) = (e,b^{-1}xb) \in K_1$ and $$b^{-1}K_1 b = K_1$$. Similar relations hold for $$K_2$$. As $$K$$ is supposed to be nontrivial, we have $$K_1 \neq K_2$$.

The exchange automorphism $$\psi : (x,y) \mapsto (y,x)$$ exchanges the subgroup $$K_1$$ and $$K_2$$. Thus, neither $$K_1$$ nor $$K_2$$ is invariant under all the automorphisms, so neither is characteristic. Therefore, $$K_1$$ and $$K_2$$ are both normal subgroups of $$G$$ that are not characteristic.

When $$K = \mathbb Z_2$$ is the cyclic group of order two, $$G = \mathbb Z_2 \times \mathbb Z_2$$ is the Klein four-group. In particular, this gives a counterexample where the ambient group is an abelian group.

### Example on the additive group $$\mathbb Q$$

Consider the additive group $$(\mathbb Q,+)$$ of rational numbers. The map $$\varphi : x \mapsto x/2$$ is an automorphism. As $$(\mathbb Q,+)$$ is abelian, all subgroups are normal. However, the subgroup $$\mathbb Z$$ is not sent into itself by $$\varphi$$ as $$\varphi(1) = 1/ 2 \notin \mathbb Z$$. Hence $$\mathbb Z$$ is not a characteristic subgroup.