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. 0,1in{mathbb F}.
  4. 0ne1.
  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 ztimes w, zcdot w, and zw indistinctly:

    1. (Commutativity of addition). For all z,win{mathbb F}, we have that z+w=w+z.
    2. (Commutativity of multiplication). Similarly, zw=wz for all z,win{mathbb F}.
    3. (Associativity of addition). For all z,w,vin{mathbb F}, we have that z+(w+v)=(z+w)+v.
    4. (Associativity of multiplication). Similarly, z(wv)=(zw)v for all z,w,vin{mathbb F}.
    5. (Distributivity). For all z,v,win{mathbb F}, one has that w(z+v)=wz+wv.
    6. (Additive identity). z+0=z for all zin{mathbb F}.
    7. (Multiplicative identity). z 1=z for all zin{mathbb F}.
    8. (Additive inverses). For any zin{mathbb F} there is a win{mathbb F} such that z+w=0.
    9. (Multiplicative inverses). For any zin{mathbb F}, if zne0, there is a win{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 ztimes0=0 for any zin{mathbb F}.

    Proof. First note that 0+0=0 by property 6. Given zin{mathbb F}, we have z0=z(0+0)=z0+z0, by distributivity. If we know that (for any t,win{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 zin{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 1times0=0 by Lemma 2. But 1ne0. {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, simsubseteq Xtimes X, such that:

    1. xsim x for all xin X. (sim is reflexive.)
    2. For any x,yin X, if xsim y then ysim x. (sim is symmetric.)
    3. For any x,y,zin X, if xsim y and ysim z, then xsim z. (sim is transitive.)

    Given a set X, an equivalence relation sim on X, and an element xin X, the equivalence class of x is the set of all members of X that are sim-related (equivalent) to x, {}[x]_sim ={yin X: xsim 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 : xin 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, mequiv kmod n, iff n|(m-k).

    It is easy to check that mequiv kmod 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 xsim y iff xequiv ymod 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]_ntimes[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 xequiv ymod n, and zequiv wmod n, then also x+zequiv y+wmod n.
    2. If xequiv ymod n, and zequiv wmod n, then also xzequiv ywmod 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.

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