Category Archives: Algebra

Algebra

A prime ideal that is not a maximal ideal

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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.

Algebra

Four elements rings

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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.

Algebra

Group homomorphism versus ring homomorphism

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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.

Algebra

A group G isomorph to the product group G x G

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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 \(G \times G\).

Algebra

Isomorphism of factors does not imply isomorphism of quotient groups

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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 \[
G / H \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \ncong \mathbb{Z}_4 \cong G / K.\]

Algebra

A Commutative Ring with Infinitely Many Units

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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 \(u \gt 1\).

Algebra

A normal extension of a normal extension may not be normal

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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.

Algebra

The image of an ideal may not be an ideal

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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 \(2x \notin 2\mathbb Z\).

Algebra

A group that is not a semi-direct product

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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.

Algebra

A normal subgroup that is not a characteristic

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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.