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-   -   Basic Number Theory 19: Introducing rings and fields (https://www.mersenneforum.org/showthread.php?t=22189)

 Nick 2017-04-11 18:37

Basic Number Theory 19: Introducing rings and fields

Modern Number Theory makes extensive use of abstract algebraic concepts.
We have already taken a brief look at the idea of a group.
Here we take a first look at rings and fields.

[U]Definitions[/U]
A [B]ring[/B] consists of a set $$R$$ (of elements - it doesn't matter what they are) together with functions $$+,\cdot:R\times R\rightarrow R$$
(i.e. each function combines 2 elements of $$R$$ in some way to form a new element of $$R$$)
which satisfy the following conditions.
For any elements $$x,y\in R$$, we write $$+(x,y)$$ as $$x+y$$ and $$\cdot(x,y)$$ as $$x\cdot y$$ or simply $$xy$$, just as with ordinary addition and multiplication.
The conditions are known as the ring [B]axioms[/B], and they are as follows.

1. Addition is associative, i.e. for all $$x,y,z\in R$$ we have (x+y)+z=x+(y+z).
2. The ring has a neutral element under addition, which we write as 0. In other words, for all $$x\in R$$ we have $$x+0=x$$ and $$0+x=x$$.
3. Every element $$x\in R$$ has an inverse under addition, which we write as $$-x$$. Thus, for all $$x\in R$$, $$x+(-x)=0$$ and $$(-x)+x=0$$.
4. Addition is commutative: for all $$x,y\in R$$, $$x+y=y+x$$.

RING AXIOMS FOR MULTIPLICATION
5. Multiplication is associative, i.e. for all $$x,y,z\in R$$ we have (xy)z=x(yz).
6. The ring has a neutral element under multiplication, which we write as 1. In other words, for all $$x\in R$$ we have $$x\cdot 1=x$$ and $$1\cdot x=x$$.

7. Multiplication is distributive over addition, i.e. for all $$x,y,z\in R$$, $$x(y+z)=xy+xz$$ and $$(x+y)z=xz+yz$$.
(Just as with ordinary numbers, we agree that multiplication has a higher priority than addition, so $$xy+xz$$ means $$(xy)+(xz)$$, for example.
This reduces the number of brackets we need to use, making everything easier to read.)

A ring $$R$$ (with its addition and multiplication functions) is called [B]commutative[/B] if it also satisfies:
8. Multiplication is commutative: for all $$x,y\in R$$, $$xy=yx$$.

An [B]integral domain[/B] (or simply a [B]domain[/B]) is a commutative ring $$R$$ with the following 2 properties as well:
9. $$1\neq 0$$ (i.e. the neutral elements under addition and multiplication are two distinct elements of the ring).
10a. The ring has no zero divisors. This means that, for all $$x,y\in R$$, if $$xy=0$$ then $$x=0$$ or $$y=0$$.

A [B]field[/B] is a commutative ring $$R$$ with these 2 properties too:
9. $$1\neq 0$$ (as above).
10b. Each non-zero element of the ring has an inverse under multiplication, i.e. for all $$x\in R$$ with $$x\neq 0$$, there exists an element which we write as $$x^{-1}\in R$$ such that $$xx^{-1}=1$$ and $$x^{-1}x=1$$.

We shall prove shortly that every field is an integral domain.

[U]Examples[/U]
The set $$\mathbb{Z}$$ of all integers (with ordinary addition and multiplication) is an integral domain but not a field.
The set $$\mathbb{Q}$$ of all rational numbers (again with ordinary + and $$\cdot$$) is a field.
The same holds for the set $$\mathbb{R}$$ of all real numbers and the set $$\mathbb{C}$$ of all complex numbers.
The set $$\mathbb{N}$$ of all natural numbers is not even a ring.

The [B]zero ring[/B] is the ring $$R=\{0\}$$ with $$0+0=0$$ and $$0\cdot 0=0$$ (and therefore $$1=0$$).
All elements of this ring are equal to each other (since there is only one) so all axioms in the above list ending with an equation are automatically satisfied.
It follows immediately that the zero ring is a commutative ring (but not an integral domain).

For any prime number $$p$$, the set $$\mathbb{Z}/p\mathbb{Z}$$ of all integers modulo $$p$$ (taken with addition and multiplication modulo $$p$$) is a field (see proposition 32).
For any integer $$n>1$$ that is not prime, $$\mathbb{Z}/n\mathbb{Z}$$ is a commutative ring but not an integral domain (as it has zero divisors).

The set $$\mathbb{Z}[i]$$ of all Gaussian integers is an integral domain (but not a field).

A 2x2 [B]matrix[/B] over $$\mathbb{R}$$ is an ordered pair of ordered pairs of real numbers, which we write as a square block, e.g.
$\left( \begin{array}{cc} 2 & -7 \\ \frac{1}{2} & \pi \end{array} \right)$
We write $$M_2(\mathbb{R})$$ for the set of all 2x2 matrices over $$\mathbb{R}$$.
$\left( \begin{array}{cc} a & b \\ c & d \end{array} \right) +\left( \begin{array}{cc} a' & b' \\ c' & d' \end{array} \right) =\left( \begin{array}{cc} a+a' & b+b' \\ c+c' & d+d' \end{array} \right)$
and matrix multiplication by
$\left( \begin{array}{cc} a & b \\ c & d \end{array} \right) \left( \begin{array}{cc} a' & b' \\ c' & d' \end{array} \right) =\left( \begin{array}{cc} aa'+bc' & ab'+bd' \\ ca'+dc' & cb'+dd' \end{array} \right)$
Under these operations, $$M_2(\mathbb{R})$$ is a ring but it is not commutative,
e.g. if $$A=\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array} \right)$$ and $$B=\left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)$$ then $$AB=\left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)$$ but $$BA=\left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right)$$
so $$AB\neq BA$$.
The zero element in the ring $$M_2(\mathbb{R})$$ is $$\left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right)$$ while the 1 is $$\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)$$.

[U]Basic properties of ring elements[/U]
Let $$R$$ be a ring (any ring).
For any integer $$n\geq 3$$ and any elements $$a_1,a_2,\ldots,a_n\in R$$, the expression $$a_1+a_2+\ldots +a_n$$ has the same value no matter where we place the brackets by propostion 70,
and the same holds for the product $$a_1a_2\ldots a_n$$.
The neutral elements $$0\in R$$ and $$1\in R$$ are unique, by proposition 71.
And, for any $$x\in R$$, its additive inverse $$-x$$ is unique, as is its multiplicative inverse $$x^{-1}$$ if it exists, by proposition 72.

In fact, the ring axioms for addition can be summarized by saying that, under addition, $$R$$ is an abelian group.
So all the propositions which we have already proved for groups apply immediately.
The only difference is in notation: for groups in general, we wrote the binary operation as multiplication (here we write it as addition), the neutral element as 1 (here it is 0)
and the inverse of an element $$x$$ as $$x^{-1}$$ (here it is $$-x$$).
For example, for any $$x,y,z\in R$$, it follows from proposition 73 that if $$x+z=y+z$$ or $$z+x=z+y$$ then $$x=y$$ (and similarly if $$x+z=z+y$$ as addition here is commutative).

What is new for rings is the interaction between addition and multiplication.
For any number $$x$$, we know that the product of $$x$$ and 0 is 0.
To prove that this is true in other rings as well, we need to show that it follows from the axioms, but there are so many that it is difficult to know where to start.
However, as the special property of 0 concerns addition but we are using it here with multiplication, we must use the only axiom linking the two.
With that insight, the proof becomes easy.

[U]Proposition 107[/U]
Let $$R$$ be a ring.
Then (i) for all $$x\in R$$, $$x\cdot 0=0$$ and $$0\cdot x=0$$;
and (ii) for all $$x,y\in R$$, $$(-x)y=-(xy)=x(-y)$$.

[U]proof[/U]
(i) For any $$x\in R$$, $$x(0+0)=x\cdot 0+x\cdot 0$$ by axiom (7).
But $$0+0=0$$ by axiom (2) so $$x(0+0)=x\cdot 0$$ as well, and therefore $$x\cdot 0+x\cdot 0=x\cdot 0$$.
By the cancellation law (proposition 73 but written additively) it follows that $$x\cdot 0=0$$.
Similarly, $$0\cdot x=(0+0)x=0\cdot x+0\cdot x$$ so $$0\cdot x=0$$.

(ii) Take any $$x,y\in R$$.
Then $$xy+(-x)y=(x+(-x))y$$ by axiom (7) and $$x+(-x)=0$$ by axiom (3) so $$xy+(-x)y=0\cdot y=0$$ by part (i) above.
And $$xy+(-(xy))=0$$ too by axiom (3) so $$xy+(-x)y=xy+(-(xy))$$ hence, by the cancellation law again, $$(-x)y=-(xy)$$.
Similarly, $$xy+x(-y)=x(y+(-y))=x\cdot 0=0=xy+(-(xy))$$ and cancelling gives $$x(-y)=-(xy)$$. ∎

Thus, for any elements $$x,y$$ of a ring, we may simply write $$-xy$$ without ambiguity, and we abbreviate $$x+(-y)$$ to $$x-y$$.

We have a cancellation law for addition, but with multiplication we need to be a little more careful.

[U]Proposition 108[/U]
Let $$R$$ be an integral domain.
Then, for all $$x,y,z\in R$$, if $$xz=yz$$ and $$z\neq 0$$ then $$x=y$$.

[U]proof[/U]
Take any $$x,y,z\in R$$.
If $$xz=yz$$ then $$xz-yz=yz-yz=0$$ by axiom (3) and $$xz-yz=(x-y)z$$ by axiom (7) so $$(x-y)z=0$$.
As $$R$$ is an integral domain and $$z\neq 0$$, it follows that $$x-y=0$$ by axiom (10a).
And $$y-y=0$$ by axiom (3), so $$x-y=y-y$$ hence $$x=y$$ by the cancellation law for addition. ∎

The condition that $$R$$ is an integral domain in the above proposition is necessary.
For example, in the integers modulo 6, we have $$\bar{3}\times\bar{3}=\bar{3}\times\bar{1}$$ but cannot conclude that $$\bar{3}=\bar{1}$$.

Let $$R$$ be a ring. We call $$a\in R$$ a [B]unit[/B] if there exists $$b\in R$$ such that $$ab=1$$ and $$ba=1$$.
We write $$R^*$$ for the set of all units in $$R$$.
(This is consistent with our earlier definition, just allowing for the case where multiplication is not commutative.)

A [B]subring[/B] of a ring $$R$$ is a subset $$S\subset R$$ which itself forms a ring under the same addition and multiplication (restricted to $$S$$) and with the same 1
(they have the same 0 too by proposition 77).
A [B]subfield[/B] of a field $$F$$ is a subset $$G\subset F$$ which itself forms a field under the same addition and multiplication (restricted to $$G$$).

[U]Examples[/U]
[LIST][*]$$\mathbb{Q}$$ is a subfield of $$\mathbb{R}$$ and $$\mathbb{Z}$$ is a subring of $$\mathbb{R}$$.[*]$$\{a+2bi:a,b\in\mathbb{Z}\}$$ is a subring of $$\mathbb{Z}[i]$$.[*]$$\{0\}$$ is not a subring of $$\mathbb{Z}$$.[*]$$\left\{\left( \begin{array}{cc} a & b \\ 0 & d \end{array} \right):a,b,d\in\mathbb{R}\right\}$$ is a subring of $$M_2(\mathbb{R})$$.[/LIST][U]Proposition 109[/U]
Let $$R$$ be a ring and $$S$$ a subset of $$R$$ which contains the element 1 and is closed under subtraction and multiplication.
Then $$S$$ is a subring of $$R$$.

[U]proof[/U]
Under addition, $$S$$ is a subgroup of $$R$$ by proposition 78.
Under multiplication, $$S$$ is closed and contains the 1 of $$R$$ by assumption, and axioms (5) and (7) hold in $$S$$ because they hold in $$R$$.
Hence $$S$$ is a ring under the same operations and with the same 1. ∎

[U]Proposition 110[/U]
All subrings of fields are integral domains.

[U]proof[/U]
Let $$F$$ be a field and $$R\subset F$$ a subring of $$F$$.
Then $$R$$ is a commutative ring with $$1\neq 0$$.
Take any $$x,y\in R$$ and suppose that $$xy=0$$ but $$y\neq 0$$.
Then, working in $$F$$, $$y$$ has a multiplicative inverse $$y^{-1}$$ and
$x=x\cdot 1=x(yy^{-1})=(xy)y^{-1}=0\cdot y^{-1}=0$
by proposition 107(i). Hence $$R$$ is an integral domain. ∎

In particular, a field is a subring of itself, so every field is an integral domain.

The converse of proposition 110 also holds: every integral domain is a subring of some field.
To prove this, we must construct the field from the domain.
Let $$R$$ be an integral domain and define a relation $$\sim$$ on the set $$R\times (R\setminus\{0\})$$
(i.e. the set of all ordered pairs of elements of $$R$$ with the 2nd element non-zero) as follows:
$(a,b)\sim (c,d)\Leftrightarrow ad=cb.$
Take any $$a,b,c,d,e,f\in R$$ with $$b,d,f$$ non-zero.
Then $$ab=ab$$ so $$(a,b)\sim (a,b)$$ and $$\sim$$ is reflexive.
If $$(a,b)\sim (c,d)$$ then $$ad=cb$$ so $$cb=ad$$ and therefore $$(c,d)\sim (a,b)$$, making $$\sim$$ symmetric.
Suppose $$(a,b)\sim (c,d)$$ and $$(c,d)\sim (e,f)$$.
Then $$ad=cb$$ so $$adf=bcf$$, and $$cf=ed$$ so $$bcf=bed$$.
Thus $$adf=bed$$ with $$d\neq 0$$ and $$R$$ is an integral domain so $$af=eb$$, hence $$(a,b)\sim (e,f)$$, making $$\sim$$ transitive.
It follows that $$\sim$$ is an equivalence relation.
For any $$a,b\in R$$ with $$b\neq 0$$, we shall write $$[a,b]$$ for the equivalence class containing $$(a,b)$$ under the relation $$\sim$$,
and we write $$Q(R)$$ for the set of all the equivalence classes.
We define addition and multiplication on $$Q(R)$$ as follows: for any $$a,b,c,d\in R$$ with $$b\neq 0\neq d$$,
$[a,b]+[c,d]=[ad+bc,bd]\mbox{ and }[a,b][c,d]=[ac,bd].$
We must check that this is well-defined.
Firstly, as $$R$$ is an integral domain, $$bd\neq 0$$ so $$[ad+bc,bd]$$ and $$[ac,bd]$$ are valid elements of $$Q(R)$$.
Take any $$a',b',c',d'\in R$$ with $$b'\neq 0\neq d'$$ and suppose that $$(a',b')\sim (a,b)$$ and $$(c',d')\sim (c,d)$$.
Then $$ab'=a'b$$ and $$cd'=c'd$$
so $$(ad+bc)b'd'=(a'd+b'c)bd'=(a'd'+b'c')bd$$ and $$acb'd'=a'c'bd$$
hence $$(ad+bc,bd)\sim (a'd'+b'c',b'd')$$ and $$(ac,bd)\sim (a'c',b'd')$$.
Thus the definitions are independent of the elements chosen to represent their equivalence classes, making addition and multiplication on $$Q(R)$$ well-defined.

[U]Proposition 111[/U]
Let $$R$$ be an integral domain.
Then $$Q(R)$$ with addition and multiplication as defined above is a field.

[U]proof[/U]
Axiom 1: for all $$[a,b],[c,d],[e,f]\in Q(R)$$,
$\begin{eqnarray*} ([a,b]+[c,d])+[e,f] & = & [ad+bc,bd]+[e,f] = [(ad+bc)f+bde,bdf] \\ & = & [adf+b(cf+de),bdf] = [a,b]+[cf+de,df] \\ & = & [a,b]+([c,d]+[e,f]). \end{eqnarray*}$
Axiom 2: for all $$[a,b]\in Q(R)$$, $$[a,b]+[0,1]=[a\cdot 1+b\cdot 0,b\cdot 1]=[a,b]$$ and similarly $$[0,1]+[a,b]=[a,b]$$.
Axiom 3: for all $$[a,b]\in Q(R)$$, $$[a,b]+[-a,b]=[ab-ab,b^2]=[0,b^2]=[0,1]$$ and similarly $$[-a,b]+[a,b]=[0,1]$$.
Axiom 4: for all $$[a,b],[c,d]\in Q(R)$$, $$[a,b]+[c,d]=[ad+bc,bd]=[cb+da,db]=[c,d]+[a,b]$$.
Axiom 5: to multiply in $$Q(R)$$, we multiply the corresponding coordinates.
And multiplication is associative in $$R$$, so it is in $$Q(R)$$, too.
Axiom 6: the neutral element of $$Q(R)$$ under multiplication is $$[1,1]$$.
Axiom 7: for all $$[a,b],[c,d],[e,f]\in Q(R)$$,
$\begin{eqnarray*} [a,b]([c,d]+[e,f]) & = & [a,b][cf+de,df] = [a(cf+de),bdf] \\ & = & [acbf+bdae,bdbf] = [ac,bd]+[ae,bf] \\ & = & [a,b][c,d]+[a,b][e,f]) \end{eqnarray*}$
and similarly $$([a,b]+[c,d])[e,f]=[a,b][e,f]+[c,d][e,f]$$.
Axiom 8: multiplication in $$R$$ is commutative, so it is in $$Q(R)$$, too.
Axiom 9: we have $$(1,1)\not\sim (0,1)$$ since $$1\neq 0$$ in $$R$$.
Axiom 10b: for any $$[a,b]\in Q(R)$$, if $$(a,b)\not\sim (0,1)$$ then $$a\neq 0$$ so $$[b,a]\in Q(R)$$ with $$[a,b][b,a]=[ab,ba]=[1,1]$$ and $$[b,a][a,b]=[1,1]$$. ∎

We call $$Q(R)$$ the [B]quotient field[/B] or [B]field of fractions[/B] of $$R$$.
For any $$a,b\in R$$, we have $$[a,1]+[b,1]=[a+b,1]$$ and $$[a,1][b,1]=[ab,1]$$.
Also, the neutral elements of $$Q(R)$$ are $$[0,1]$$ under addition and $$[1,1]$$ under multiplication.
So the elements of $$Q(R)$$ which can be written with 2nd coordinate 1 behave exactly like the corresponding elements of $$R$$: $$Q(R)$$ contains a copy of $$R$$, and we may identify it with $$R$$.
For each $$a\in R$$, we write $$[a,1]$$ simply as $$a$$, and for any $$a,b\in R$$ with $$b\neq 0$$ we write $$[a,b]$$ as a fraction $$\frac{a}{b}$$.
In this way, $$R$$ becomes a subring of $$Q(R)$$.

[U]Example[/U]
$$\mathbb{Q}=Q(\mathbb{Z})$$ and we identify each integer $$n$$ with the rational number $$\frac{n}{1}$$, making $$\mathbb{Z}$$ a subring of $$\mathbb{Q}$$.

[U]Exercises[/U]
87. Show that every finite integral domain is a field.
88. Show that the zero ring is the only ring in which the 1 and the 0 are equal.
89. Let $$R$$ be a commutative ring. Prove that $$R$$ is a field if and only if $$R^*=R\setminus\{0\}$$.
89. Prove for all $$a,b,c,d\in\mathbb{R}$$ that $$\left(\begin{array}{cc}a & b \\ c & d \end{array}\right)\in M_2(\mathbb{R})^*$$ if and only if $$ad-bc\neq 0$$.
90. Let $$R=\{a+b\sqrt{2}:a,b\in\mathbb{Q}\}$$. Show that $$R$$ is a subring of $$\mathbb{R}$$. Is it also a field?

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