**Definition 1. **Let be a set. We say that the quintuple is a **field **iff the following conditions hold:

- (We say that is
**closed**under addition and multiplication.) - Properties 1–9 of the Theorem from last lecture hold with elements of in the place of complex numbers, in the place of and in the place of

The Theorem from last lecture can be restated as saying that is a field. For ease, I rewrite properties 1–9 below; as usual, we write and indistinctly:

- (Commutativity of addition). For all we have that
- (Commutativity of multiplication). Similarly, for all
- (Associativity of addition). For all we have that
- (Associativity of multiplication). Similarly, for all
- (Distributivity). For all one has that
- (Additive identity). for all
- (Multiplicative identity). for all
- (Additive inverses). For any there is a such that
- (Multiplicative inverses). For any , if there is a such that

(We will typically abuse notation and say simply that 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 is indeed some kind of multiplication, but in general they are just some abstract functions that satisfies the properties required above.

**Examples. ****1.** is *not *a field, because does not have additive inverses of any number different from

**2. **Trying to fix example 1, we could add the additive inverses to , and now we have This is *not* a field either, because does not have multiplicative inverses of any nonzero number different from

**3.** Trying to fix example 2, we could now add the multiplicative inverses to This new set is not even closed under addition or multiplication. For example, 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 This *is* a field.

**4.** is a field.

**5. ** where is addition modulo 2 (*exclusive or*) and 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 is a field, then for any

**Proof. **First note that by property 6. Given we have by distributivity. If we know that (for any ) whenever then in fact then we are done, by taking and We prove this property in Lemma 3.

Lemma 3.If is a field, then implies for any

**Proof.** Suppose that Let be an additive inverse of so By commutativity, Then But also It follows that as we wanted.

Corollary 4.If is a field, then and are different functions.

**Proof.** and by Lemma 2. But

Now we will try to generalize Example 5 above.

**Example.** **6. **Let be a positive integer. We want to define the set of numbers modulo . For this, we begin with and define an equivalence relation on it.

**Definition 5. **Given a set an **equivalence relation** on is a set of ordered pairs of elements of such that:

- for all . ( is
**reflexive**.) - For any if then . ( is
**symmetric**.) - For any if and then ( is
**transitive**.)

Given a set an equivalence relation on and an element the **equivalence class** of is the set of all members of that are -related (equivalent) to

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

Suppose that is a set and is an equivalence relation on . The **quotient set** is the collection of equivalence classes determined by

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

Now we continue with Example 6 by defining a particular equivalence relation on

**Definition 6.** Given a positive integer two integers and are said to be congruent mod in symbols, iff

It is easy to check that iff the remainder of dividing by is the same as the remainder of dividing by

Lemma 7.The relation on integers given by iff is an equivalence relation.

We write for the collection of equivalence classes of the equivalence relation of congruence mod We denote by the equivalence class of and sometimes also write for the quotient space

We are interested to see whether with the natural operations of addition and multiplication and the natural versions of and is a field. These `natural operations’ are defined as follows:

**Definition 8. **Given a positive number **addition mod ** is defined by setting

**Multiplication mod ** is defined by setting

The of is and the of is

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 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 in this case) is **well defined**.

Lemma 9.The operations and mod are well-defined. More precisely,

- If and then also
- If and then also

**Proof.** For 1., note that and for 2., note that The result follows easily from these equalities.

We are finally ready to examine Example 6: Is a field? We will study this question next lecture.

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