## 305 -4. Fields.

Definition 1. Let ${\mathbb F}$ be a set. We say that the quintuple $({\mathbb F},+,\times,0,1)$ is a field iff the following conditions hold:

1. $+:{\mathbb F}\times {\mathbb F}\to {\mathbb F}.$
2. $\times:{\mathbb F}\times {\mathbb F}\to {\mathbb F.}$ (We say that ${\mathbb F}$ is closed under addition and multiplication.)
3. $0,1\in{\mathbb F}.$
4. $0\ne1.$
5. Properties 1–9 of the Theorem from last lecture hold with elements of ${\mathbb F}$ in the place of complex numbers, ${}0$ in the place of $\hat0,$ and ${}1$ in the place of $\hat1.$

The Theorem from last lecture can be restated as saying that $({\mathbb C},+,\times,0,1)$ is a field. For ease, I rewrite properties 1–9 below; as usual, we write $z\times w,$ $z\cdot w,$ and $zw$ indistinctly:

1. (Commutativity of addition). For all $z,w\in{\mathbb F},$ we have that $z+w=w+z.$
2. (Commutativity of multiplication). Similarly, $zw=wz$ for all $z,w\in{\mathbb F}.$
3. (Associativity of addition). For all $z,w,v\in{\mathbb F},$ we have that $z+(w+v)=(z+w)+v.$
4. (Associativity of multiplication). Similarly, $z(wv)=(zw)v$ for all $z,w,v\in{\mathbb F}.$
5. (Distributivity). For all $z,v,w\in{\mathbb F},$ one has that $w(z+v)=wz+wv.$
6. (Additive identity). $z+0=z$ for all $z\in{\mathbb F}.$
7. (Multiplicative identity). $z 1=z$ for all $z\in{\mathbb F}.$
8. (Additive inverses). For any $z\in{\mathbb F}$ there is a $w\in{\mathbb F}$ such that $z+w=0.$
9. (Multiplicative inverses). For any $z\in{\mathbb F}$, if $z\ne0,$ there is a $w\in{\mathbb F}$ such that $zw=1.$

(We will typically abuse notation and say simply that ${\mathbb F}$ is a field, although it is understood that, formally, we mean the quintuple.)

In many natural cases, it will be clear that $+$ is indeed some kind of addition and $\times$ is indeed some kind of multiplication, but in general they are just some abstract functions that satisfies the properties required above.

Examples. 1. $({\mathbb N},+,\times,0,1)$ is not a field, because ${\mathbb N}$ does not have additive inverses of any number different from ${}0.$

2. Trying to fix example 1, we could add the additive inverses to ${\mathbb N}$, and now we have $({\mathbb Z},+,\times,0,1).$ This is not  a field either, because ${\mathbb Z}$ does not have multiplicative inverses of any nonzero number different from ${}1.$

3. Trying to fix example 2, we could now add the multiplicative inverses to ${\mathbb Z}.$ This new set is not even closed under addition or multiplication. For example, $3+\frac12$ is neither an integer nor the inverse of an integer. To solve this problem, lets take instead all the possible products of integers and their inverses. We now have $({\mathbb Q},+,\times,0,1).$ This is a field.

4. $({\mathbb R},+,\times,0,1)$ is a field.

5. $(\{0,1\},+,\times,0,1),$ where $+$ is addition modulo 2 (exclusive or) and $\times$ is multiplication mod 2 (and). This is a field. In a sense, it is as simple a field as we can get, since condition 4 above requires that any field has at least 2 elements.

Before looking for any more examples, let’s verify a few basic properties of fields:

Lemma 2. If ${\mathbb F}$ is a field, then $z\times0=0$ for any $z\in{\mathbb F}.$

Proof. First note that $0+0=0$ by property 6. Given $z\in{\mathbb F},$ we have $z0=z(0+0)=z0+z0,$ by distributivity. If we know that (for any $t,w\in{\mathbb F}$) whenever $t+w=t$ then in fact $w=0,$ then we are done, by taking $t=z0$ and $w=z0.$ We prove this property in Lemma 3. ${\sf QED}$

Lemma 3. If ${\mathbb F}$ is a field, then $z+w=z$ implies $w=0$ for any $z\in{\mathbb F}.$

Proof. Suppose that $z+w=z.$ Let $p$ be an additive inverse of $z,$ so $z+p=0.$ By commutativity, $p+z=0.$ Then $p+(z+w)=p+z=0.$ But also $p+(z+w)=(p+z)+w=0+w=w+0=w.$ It follows that $w=0,$ as we wanted. ${\sf QED}$

Corollary 4. If ${\mathbb F}$ is a field, then $+$ and $\times$ are different functions.

Proof. $1+0=1$ and $1\times0=0$ by Lemma 2. But $1\ne0.$ ${\sf QED}$

Now we will try to generalize Example 5 above.

Example. 6. Let $n$ be a positive integer. We want to define ${\mathbb Z}_n,$ the set of numbers modulo $n$. For this, we begin with ${\mathbb Z}$ and define an equivalence relation on it.

Definition 5. Given a set $X,$ an equivalence relation $\sim$ on $X$ is a set of ordered pairs of elements of $X,$ $\sim\subseteq X\times X,$ such that:

1. $x\sim x$ for all $x\in X$. ($\sim$ is reflexive.)
2. For any $x,y\in X,$ if $x\sim y$ then $y\sim x$. ($\sim$ is symmetric.)
3. For any $x,y,z\in X,$ if $x\sim y$ and $y\sim z,$ then $x\sim z.$ ($\sim$ is transitive.)

Given a set $X,$ an equivalence relation $\sim$ on $X,$ and an element $x\in X,$ the equivalence class of $x$ is the set of all members of $X$ that are $\sim$-related (equivalent) to $x,$ ${}[x]_\sim =\{y\in X:$ $x\sim y\}.$

The typical example of an equivalence relation is equality. Any equivalence relation is like’ equality in a sense:

Suppose that $X$ is a set and $\sim$ is an equivalence relation on $X$. The quotient set $X/\sim$ is the collection $\{[x]_\sim : x\in X\}$ of equivalence classes determined by $\sim.$

We can think of this as looking at the set $X$ from a distance. Then we cannot distinguish between points in the same equivalence class, and all we see is $X/\sim.$ In this sense, we now have equality in place of $\sim.$ Once we approach the set, we then see that what we thought were individual points are actually collections of elements of $X,$ namely, the equivalence classes.

Now we continue with Example 6 by defining a particular equivalence relation on ${\mathbb Z}:$

Definition 6. Given a positive integer $n,$ two integers $m$ and $k$ are said to be congruent mod $n,$ in symbols, $m\equiv k\mod n,$ iff $n|(m-k).$

It is easy to check that $m\equiv k\mod n$ iff the remainder of dividing $m$ by $n$ is the same as the remainder of dividing $k$ by $n.$

Lemma 7. The relation on integers given by $x\sim y$ iff $x\equiv y\mod n$ is an equivalence relation. $\Box$

We write ${\mathbb Z}_n$ for the collection of equivalence classes of the equivalence relation of congruence mod $n.$ We denote by ${}[x]_n$ the equivalence class of $x,$ and sometimes also write ${\mathbb Z}/n{\mathbb Z}$ for the quotient space ${\mathbb Z}_n.$

We are interested to see whether ${\mathbb Z}_n$ with the natural operations of addition and multiplication and the natural versions of ${}0$ and ${}1$ is a field. These natural operations’ are defined as follows:

Definition 8. Given a positive number $n,$ addition mod $n$ is defined by setting ${}[x]_n+[y]_n=[x+y]_n.$

Multiplication mod $n$ is defined by setting ${}[x]_n\times[y]_n=[xy]_n.$

The ${}0$ of ${\mathbb Z}_n$ is ${}[0]_n$ and the ${}1$ of ${\mathbb Z}_n$ is ${}[1]_n.$

There is an obstacle we need to pass in order to make sense of this definition.  Namely, $+$ is defined on the classes, but the definition depends on representatives of the classes. In principle, it could be that if we chose different representatives, we would obtain at the end a different class. The same problem occurs with the definition of $\times.$ When defining an operation on equivalence classes by looking at representatives of these classes, if it is indeed the case that no matter what representatives we choose we get at the end the same class, we say that the operation ($+$ or $\times$ in this case) is well defined.

Lemma 9. The operations $+$ and $\times$ mod $n$ are well-defined. More precisely,

1. If $x\equiv y\mod n,$ and $z\equiv w\mod n,$ then also $x+z\equiv y+w\mod n.$
2. If $x\equiv y\mod n,$ and $z\equiv w\mod n,$ then also $xz\equiv yw\mod n.$

Proof. For 1., note that $(x+z)-(y+w)=(x-y)+(z-w)$ and for 2., note that $xz-yw=xz-yz+yz-yw$ $=(x-y)z+y(z-w).$ The result follows easily from these equalities. ${\sf QED}$

We are finally ready to examine Example 6: Is $({\mathbb Z}_n,+,\times,[0]_n,[1]_n)$ a field? We will study this question next lecture.