In a ring $R$ a unit is any element $u$ that has a multiplicative inverse $v$, i.e. an element $v$ such that

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 $n \in \mathbb N$

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 $u \gt 1$.

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 $2x \notin 2\mathbb Z$.

Let’s recall that a set $R$ equipped with two operations $(R,+,\cdot)$ is a ring if and only if $(R,+)$ is an abelian group, multiplication $\cdot$ is associative and has a multiplicative identity $1$ and multiplication is left and right distributive with respect to addition.

$(\mathbb Z, +, \cdot)$ is a well known infinite ring which is commutative. The rational, real and complex numbers are other infinite commutative rings. Those are in fact fields as every non-zero element have a multiplicative inverse.

For a field $F$ (finite or infinite), the polynomial ring $F[X]$ is another example of infinite commutative ring.

Also for $n$ integer, the integers modulo n is a finite ring that is commutative. Finally, according to Wedderburn theorem every finite division ring is commutative.

So what are examples of non commutative rings? Let’s provide a couple. Continue reading Non commutative rings →

Let’s recall that a simple ring is a non-zero ring that has no two-sided ideal besides the zero ideal and itself. A division ring is a simple ring. Is the converse true? The answer is negative and we provide here a counterexample of a simple ring which is not a division ring.

We prove that for $n \ge 1$ the matrix ring $M_n(F)$ of $n \times n$ matrices over a field $F$ is simple. $M_n(F)$ is obviously not a division ring as the matrix with $1$ at position $(1,1)$ and $0$ elsewhere is not invertible.

Let’s prove first following lemma. Continue reading A simple ring which is not a division ring →