Section 4.1
Commutative rings
A commutative ring is like a vector space, except that we do not have multiplicative inverses.
A commutative ring $R$ is a set with the choice of two elements $0, 1 \in R$ and two maps $+, \cdot : R \times R \to R$ satisfying the following conditions: (Here, we write $+(a, b) = a + b$ and $\cdot (a, b) = a \cdot b$.)
- (R1)
For all $a \in R$ we have $a + 0 = 0 + a = a$.
- (R2)
For all $a, b \in R$ we have $a + b = b + a$.
- (R3)
For all $a, b, c \in R$ we have $a + (b + c) = (a + b) + c$.
- (R4)
For all $a \in R$ there exists an $(-a) \in R$ such that $a + (-a) = (-a) + a = 0$.
- (R5)
For all $a \in R$ we have $a \cdot 1 = 1 \cdot a = a$
- (R6)
For all $a, b \in R$ we have $a \cdot b = b \cdot a$.
- (R7)
For all $a, b, c \in R$ we have $a \cdot (b \cdot c) = (a \cdot b) \cdot c$.
- (R8)
For all $a, b, c \in R$ we have $a \cdot (b + c) = (a \cdot b) + (a \cdot c)$.
Here, are some standard examples.
The ring $\lbrace 0 = 1\rbrace $ with one element is a ring, with the obvious addition and multiplication maps.
Any field is a ring.
$\mathbb{Z}$ is a ring with usual addition and multiplication.
Take any ring $R$, and look at the set of polynomials
$$ R[t] = \lbrace a_0 + a_1 t + a_2 t^2 + \cdots + a_n t^n : a_0, \ldots, a_n \in R \rbrace $$
in the variable $t$ with coefficients in $R$. This is a ring with addition and multiplication given by
$$ \sum_{i}^{} a_i t^i + \sum_{i}^{} b_i t^i = \sum_{i}^{} (a_i + b_i) t^i, \quad \biggl( \sum_{i}^{} a_i t^i \biggr) \biggl( \sum_{i}^{} b_i t^i \biggr) = \sum_{i}^{} \biggl( \sum_{j+k=i}^{} a_j b_k \biggr) t^i. $$
Take any ring $R$, but now look at the set of formal power series
$$ R[[t]] = \lbrace a_0 + a_1 t + a_2 t^2 + a_3 t^3 + \cdots : a_0, a_1, \ldots \in R \rbrace $$
in the variable $t$ with coefficients in $R$. (Here, you don't worry about convergence just take the set of all series that can possibly be written down.) This is going to be a ring with addition and multiplication given similarly.
The set
$$ \mathbb{Z}/n\mathbb{Z} = \lbrace 0, 1, 2, \ldots, n-1 \rbrace $$
is a ring with addition and multiplication defined modulo $n$, i.e., taking remainder of division by $n$ after doing ordinary addition or multiplication.
For an integer $n > 0$, the set
$$ \mathbb{Z}[i] = \lbrace a + bi : a, b \in \mathbb{Z} \rbrace \subseteq \mathbb{C} $$
is a ring with usual addition and multiplication.
The set
$$ \mathbb{Z}[\tfrac{1}{2}] = \lbrace a 2^{-b} : a \in \mathbb{Z}, b \in \mathbb{Z}_{\ge 0} \rbrace \subseteq \mathbb{Q} $$
is a ring with usual addition and multiplication.
As you can see, there are lots and lots of rings, and the theory of commutative rings can be pretty delicate. Here is one concept in commutative ring theory (also called commutative algebra) that helps studying rings.
An ideal of a ring $R$ is a subset $\mathfrak{a} \subseteq R$ such that
- (I1)
$0 \in \mathfrak{a}$,
- (I2)
$a, b \in \mathfrak{a}$ implies $a + b, -a \in \mathfrak{a}$,
- (I3)
for any $a \in \mathfrak{a}$ and $r \in R$, we have $ra \in \mathfrak{a}$.
This is not the same as a subring, because (I3) is something stronger than saying that $a, b \in \mathfrak{a}$ implies $ab \in \mathfrak{a}$.
For each element $a \in R$, show that the set
$$ (a) = aR = \lbrace ra : r \in R\rbrace $$
is an ideal of $R$. An ideal that can be written in this form is called a principal ideal.
Show that the ideals of $\mathbb{Z}$ are precisely $(0), (1) = \mathbb{Z}, (2), (3), \ldots$. (Hint: take the minimal positive element in the ideal.)
A commutative ring $R$ is called a principal ideal domain, or PID for short, if
- (PID1)
$ab = 0$ implies either $a = 0$ or $b = 0$,
- (PID2)
every ideal of $R$ is principal.
So for example $\mathbb{Z}$ is a principal ideal domain.
For $k$ a field, show that the polynomial ring $k[t]$ is a principal ideal domain.
Show that $\mathbb{Z}[i]$ is a principal ideal domain. On the other hand, show that
$$ \mathbb{Z}[\sqrt{-5}] = \lbrace a + b \sqrt{-5} : a, b \in \mathbb{Z} \rbrace \subseteq \mathbb{C} $$
is not a principal ideal domain.
To satisfy number-theorists, let me define the following divisibility relation.
Let $R$ be a ring in general. For $x, y \in R$, we say that $x$ divides $y$ or write $x \mid y$ if there exists a $z \in R$ such that $xz = y$.
For instance, we always have $1 \mid x$ or $x \mid 0$ because $x = 1 \cdot x$ and $0 = x \cdot 0$. Of course, the motivation comes from $R = \mathbb{Z}$. If $R$ is a field, this relation is not very useful because for instance $x \mid y$ for arbitrary $x, y \in R \setminus \lbrace 0\rbrace $.
In a principal ideal domain, we may take the greatest common divisor by adding ideals.
Let $\mathfrak{a}, \mathfrak{b} \subseteq R$ be two ideals. Show that
$$ \mathfrak{a} + \mathfrak{b} = \lbrace a + b : a \in \mathfrak{a}, b \in \mathfrak{b} \rbrace \subseteq R $$
is again an ideal of $R$.
Assume that $R$ is a principal ideal domain. Consider two elements $a, b \in R$ so that $(a)$ and $(b)$ are two ideals of $R$. Because $(a) + (b)$ is an ideal of $R$, we may find a $d \in R$ (it is not uniquely determined!) such that
$$ (a) + (b) = (d). $$
Show that $d$ divides both $a$ and $b$, and show that if some $e \in R$ divides both $a$ and $b$, then $e$ divides $d$. Check that this agrees with the usual notion of a greatest common divisor when $R = \mathbb{Z}$.
We can quotient rings by ideals.
Let $R$ be a commutative ring and $\mathfrak{a} \subseteq R$ be an ideal. Show that
$$ R / \mathfrak{a} = R / (x \sim a + x \text{ for } x \in R, a \in \mathfrak{a}) $$
inherits a structure of a ring from $R$. (You first need to check that the relation is an equivalence relation. Then you will need to check that the addition and multiplication maps $[x] + [y] = [x+y]$ and $[x] [y] = [x y]$ are well-defined. After this, you should check the ring axioms.)
The example $\mathbb{Z}/n\mathbb{Z}$ is actually an instance of this. Because $(n) = n\mathbb{Z}$ is an ideal of $\mathbb{Z}$, the quotient $\mathbb{Z}/n\mathbb{Z}$ naturally is a ring.