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Linear Algebra Gathering and Sharpening Tools

Ryan Higginbottom

Section 2.1 Fields

In this section we will consider the real numbers and study their most important properties in a general setting. In the process, we will learn how to handle axioms and abstract algebraic concepts.
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In Chapter 1, we used the real numbers as the familiar world within which all of our calculations took place. Before we offer any definitions or results in this chapter, we will ponder exactly what properties of the real numbers were essential to these calculations. (The reader is likely so familiar with the real numbers that its important properties have been internalized and are not in the conscious mind. In this section we will make those properties explicit.)
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The most basic concept in this book is the linear equation. What properties of the real numbers are used to solve a simple linear equation? In what follows we will solve the equation \(2x+7=12\) and draw to the surface some of these important properties:
\begin{align} 2x + 7 \amp =12\tag{2.1}\\ (2x+7)-7 \amp =12-7\tag{2.2}\\ 2x + (7-7) \amp =5\tag{2.3}\\ 2x + 0 \amp =5\tag{2.4}\\ 2x \amp =5\tag{2.5}\\ \frac{1}{2}(2x) \amp = \frac{1}{2}5\tag{2.6}\\ (\frac{1}{2}2)x \amp = \frac{5}{2}\tag{2.7}\\ (1)x \amp = \frac{5}{2}\tag{2.8}\\ x \amp = \frac{5}{2}\text{.}\tag{2.9} \end{align}
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No student wants to write out all of these steps, and no instructor wants to grade such a solution! But the point is to notice the properties of the real numbers which are vital to solving an equation like this and which we usually ignore.
  • In (2.3) we used the associativity of addition in \(\rr\text{.}\) That is, we can move the parentheses around in addition and still have an equivalent expression. (The reader should see subtraction as a form of addition.)
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  • In (2.4) we used the fact that \(7\) has an additive inverse—a number we can add to \(7\) to get \(0\text{.}\) (The additive inverse of \(7\) is \(-7\text{.}\))
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  • In (2.5) we used the fact that \(0\) is an additive identity in \(\rr\)—when we add \(0\) to any real number \(r\text{,}\) we get \(r\) again.
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  • In (2.7) we used the associativity of multiplication in \(\rr\text{.}\) We can move the parentheses around in multiplication just like we can in addition.
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  • In (2.8) we used the fact that \(\frac{1}{2}\) is the multiplicative inverse of \(2\) in \(\rr\text{.}\) In other words, we can multiply \(2\) by \(\frac{1}{2}\) to get \(1\text{.}\)
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  • Finally, in (2.9) we used the fact that \(1\) is a multiplicative identity in \(\rr\)—when we multiply any real number by \(1\text{,}\) that real number is unchanged.
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By identifying these properties, our goal is to envision other mathematical realms in which solving linear equations would work the same way it does within \(\rr\text{.}\) Toward this end, we now define an algebraic object called a “field” which has all of the properties used above (plus a few we haven’t yet mentioned).
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Definition 2.1.1.

A field is a set \(\ff\) with operations \(+\) and \(\cdot\) and distinct elements \(0, 1 \in \ff\) such that all of the following properties hold.
  1. For all \(a, b \in \ff\text{,}\) \(a+b \in \ff\text{.}\) (We say that \(\ff\) is closed under addition.)
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  2. For all \(a, b \in \ff\text{,}\) \(a\cdot b \in \ff\text{.}\) (We say that \(\ff\) is closed under multiplication.)
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  3. For all \(a, b \in \ff\text{,}\) \(a+b = b+a\text{.}\) (We say that addition in \(\ff\) is commutative.)
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  4. For all \(a, b, c \in \ff\text{,}\) \(a+(b+c)=(a+b)+c\text{.}\) (We say that addition in \(\ff\) is associative.)
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  5. For each \(a \in \ff\text{,}\) \(a+0=0+a=a\text{.}\) (We say that \(0\) is an additive identity in \(\ff\text{.}\))
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  6. For each \(a \in \ff\text{,}\) there exists an element \(b \in \ff\text{,}\) such that \(a+b=b+a=0\text{.}\) (We say that each \(a\) has an additive inverse in \(\ff\text{.}\))
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  7. For each \(a, b \in \ff\text{,}\) \(a\cdot b = b\cdot a\text{.}\) (We say that multiplication is commutative in \(\ff\text{.}\))
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  8. For each \(a, b, c \in \ff\text{,}\) \(a \cdot (b \cdot c) = (a\cdot b) \cdot c\text{.}\) (We say that multiplication is associative in \(\ff\text{.}\))
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  9. For each \(a \in \ff\text{,}\) \(a \cdot 1 = 1 \cdot a = a\text{.}\) (We say that \(1\) is a multiplicative identity in \(\ff\text{.}\))
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  10. For each \(a \in \ff\) with \(a \neq 0\text{,}\) there exists an element \(c \in \ff\) such that \(a\cdot c = c \cdot a = 1\text{.}\) (We say that every nonzero element \(a\) in \(\ff\) has a multiplicative inverse.)
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  11. For each \(a, b, c \in \ff\text{,}\) \(a \cdot (b+c) = (a\cdot b) + (a\cdot c)\text{.}\) (We say that addition and multiplication in \(\ff\) satisfy the distributive law.)
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Note 2.1.2.

Axioms 5 and 6 could perhaps be stated with a bit more precision. Some texts state axiom 5 like this: the set \(\ff\) must have an additive identity, that is, an element \(e\) such that \(x + e = e + x = x\) for all \(x \in \ff\text{.}\) Those same texts may also state axiom 6 without reference to the symbol \(0\text{:}\) for each \(a \in \ff\text{,}\) there exists an element \(b \in \ff\) such that \(a + b = b + a = e\text{,}\) where \(e\) is the additive identity of the field.
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For some students, the use of the symbol \(0\) might be helpful, reminding them of the essential property of the additive identity through familiarity. But this may be confusing when handling a set which contains the symbol \(0\) when that symbol does not function as the relevant additive identity.
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Our point here is that the use of \(0\) in the axioms for a field (and some of the subsequent theorems) is intended to stand in for a generic symbol for the additive identity. Similarly, the symbol \(1\) stands in as the generic symbol for the multiplicative identity. (A reader could imagine a restatement of axioms 9 and 10 above using a letter as the multiplicative identity in the way we have demonstrated axioms 5 and 6 can be restated.)
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The reader should also note that in a generic field the term “nonzero” means “not the additive identity.”
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A note about axioms.

What are presented in Definition 2.1.1 are known as the axioms of a field. This may be the reader’s first exposure to axioms in mathematics, and this is worthy of a comment or two.
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Much of theoretical mathematics is built upon axiomatic reasoning. The thinking goes like this: If we assume a limited number of properties are true, and we assume nothing beyond those properties, what else follows necessarily? In this way we can ask what is true of a field, not just what is true of the real numbers. While the real numbers may have specific properties that a general field does not, anything that is true of a general field must be true of the real numbers.
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Working through some examples (and non-examples) will help us make sense of this definition.
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Example 2.1.3.

The set of real numbers \(\rr\) is a field. (If the real numbers were not a field, then we wouldn’t have done a very good job of abstracting the properties of the real numbers for this definition!)
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Example 2.1.4.

The complex numbers are defined in this way:
\begin{equation*} \cc = \{a + bi \mid a, b \in \rr \}\text{,} \end{equation*}
where \(i^2 = -1\text{.}\) Addition and multiplication are defined in this way:
\begin{align*} (a+bi) + (c+di) \amp =(a+c) + (b+d)i\\ (a+bi) \cdot (c+di) \amp =(ac-bd)+(ad+bc)i\text{.} \end{align*}
Notice that the addition and multiplication occuring within the parentheses on the right side of these equations are happening within \(\rr\text{.}\) In this way, showing that some of the field axioms hold for \(\cc\) depends on \(\rr\) being a field.
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The elements \(0\) and \(1\) in \(\cc\) are
\begin{align*} 0 \amp =0+0i\\ 1 \amp =1+0i\text{,} \end{align*}
and these elements are not equal. (We recall that part of the definition of a field is that the elements \(0\) and \(1\) are distinct.)
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With these definitions, one can check that \(\cc\) satisfies the properties of a field. We will prove a few of these properties, and we will assign a few of these proofs in the exercises.
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The definitions of \(+\) and \(\cdot\) in \(\cc\) show that \(\cc\) is closed under addition and multiplication. (Notably, this relies on \(\rr\) being closed under addition and multiplication!) To prove that the addition in \(\cc\) is commutative, we consider two complex numbers \(a+bi\) and \(c+di\text{.}\) Then
\begin{align} (a+bi) + (c+di) \amp = (a+c) + (b+d)i \tag{2.10}\\ \amp = (c+a) + (d+b)i\tag{2.11}\\ \amp = (c+di) + (a+bi)\text{.}\tag{2.12} \end{align}
(We note that line (2.11) used the fact that addition of real numbers is commutative.) This proves that addition in \(\cc\) is commutative.
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We will also prove that \(0\) is an additive identity in \(\cc\text{.}\) If \(a+bi \in \cc\text{,}\) then
\begin{equation*} (a+bi)+ (0+0i) = (a+0) + (b+0)i = a+bi\text{.} \end{equation*}
(This uses the fact that \(0 \in \rr\) is an additive identity.) We note that although the definition of a field requires us to examine addition by \(0\) “on both sides,” since we just proved that addition is commutative, what we have already shown is sufficient.
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Example 2.1.5.

The set of non-negative real numbers \(\rr^{\ge 0}\) is not a field. (When thinking about a subset of the real numbers, we will assume that the usual addition and multiplication are in view unless otherwise stated.) While this set is closed under addition and multiplication, it does not contain additive inverses for positive real numbers. For example, the number \(9\) has no additive inverse in this set.
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Example 2.1.6.

The set of rational numbers \(\qq\) is a field. The rational numbers are defined as the set of all quotients (hence the symbol \(\qq\)) of integers; more formally,
\begin{equation*} \qq = \left\{ \frac{a}{b} \mid a, b \in \zz, b \neq 0 \right\}\text{.} \end{equation*}
We do not need to show that all of the field axioms hold in order to prove that \(\qq\) is a field. Since \(\qq\) is a subset of \(\rr\text{,}\) and since it therefore inherits its operations from \(\rr\text{,}\) some of the field axioms hold automatically. (These would include the associativity and commutativity of addition. What else holds “by inheritance?”) Checking some of these details is left to the exercises.
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Note 2.1.7.

This notion of “inherited” properties comes up with some frequency, so we will undertake a brief explanation. Supppser we have \(A \subseteq B\) for two sets \(A\) and \(B\text{,}\) and we have an operation \(\oplus\) defined on \(B\text{.}\) Any property that is intrinsic to the operation (rather than to the set) will hold in both \(B\) and \(A\text{.}\) For example, if \(\oplus\) is commutative in \(B\text{,}\) then \(\oplus\) will also be commutative in \(A\) since \(A \subseteq B\text{.}\)
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Not all field properties are “inherited” in this same way with subsets because some properties depend on the elements of the set \(A\) and not just the operation. Example 2.1.5 is a good example of this; \(\rr^{\ge 0}\) is a subset of \(\rr\) with the same operations, but it is not a field because (for example) the property of additive inverses is not an inherited property in the same way that the commutativity or associativity of addition is inherited.
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We refer the reader to Appendix A for the definitions of \(\zz_n\) and \(\ff_p\text{,}\) related notational conventions, and a refresher on modular arithmetic. All of this is necessary for the following example.
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Example 2.1.8.

When \(p\) is a prime integer, then \(\ff_p\) is a field.
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Working within \(\ff_p\) can be a bit destabilizing at first, as the calculations take some practice. However, this effort pays off because the smallest finite fields offer some of the most tangible sandboxes in which to play. We will be using \(\ff_p\) for small \(p\) throughout this chapter to develop some interesting examples.
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Part of the definition of a field is that one can divide by any nonzero element. But because we work within \(\rr\) so often, division within \(\ff_p\) is strange. In \(\ff_3\text{,}\) for instance, the multiplicative inverse of \(2\) is \(2\text{.}\) (Another way to say this is that, within \(\ff_3\text{,}\) the element that acts the most like \(\frac{1}{2}\) is \(2\text{.}\))
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The field \(\ff_2\) contains only the elements \(0\) and \(1\text{,}\) and it therefore is a model for anything that is binary. For this reason, working in \(\ff_2\) is often very useful in computer science.
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Having defined fields, we now turn to the consequences of this definition. In other words, if a set with two operations meets the definition of a field, what else must also be true? The following theorem presents some basic results that flow from the axioms of a field. Some of these will look familiar because they are true in \(\rr\) and that’s where most people are comfortable working.
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Proof.

We will prove a few of these and leave the rest as exercises.
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To prove (Item 1) that the additive identity is unique, we must prove that \(0\) is the only element in \(\ff\) which has the properties of an additive identity. We suppose that \(a \in \ff\) is such that \(a+x=x+a=x\) for every element \(x \in \ff\text{.}\) Since this must be true for every \(x\text{,}\) it must be true for \(x=0\text{.}\) The previous equation then becomes
\begin{equation*} a+0=0+a=0\text{.} \end{equation*}
Since \(0\) is an additive identity, \(a+0=a\text{,}\) which combined with the fact that \(a+0=0\) means that \(a=0\text{.}\) This proves that the additive identity, \(0\text{,}\) is unique.
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To prove (Item 3) that additive inverses are unique, we must prove that for any element \(x \in \ff\text{,}\) there is a unique element that behaves like an additive inverse of \(x\text{.}\) We let \(x \in \ff\) and we suppose that \(y, z \in \ff\) are both additive inverses of \(x\text{.}\) We wish to show that \(y=z\text{.}\)
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Our assumptions mean that
\begin{align*} x+y \amp =y+x=0\\ x+z \amp =z+x=0\text{.} \end{align*}
We then use these assumptions, along with some of the properties of fields (the associativity of addition and the properties of the additive identity) in this calculation:
\begin{align*} x+y \amp =0\\ z+(x+y) \amp =z+0\\ (z+x)+y \amp =z\\ 0+y \amp =z\\ y \amp =z \text{.} \end{align*}
This proves that additive inverses are unique.
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We offer one final note on notation. We will often use juxtaposition to denote multiplication within a field. That is, we may write \(xy\) instead of \(x \cdot y\) to indicate the product of two elements in a field. We trust the reader will adjust quickly to this seismic shift.
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Reading Questions Reading Questions

1.

Carry out the following calculations within \(\ff_7\text{.}\) (All answers should be a number between \(0\) and \(6\text{.}\))
  1. \(\displaystyle 4 + 3\)
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  2. \(\displaystyle 5 \cdot 6\)
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  3. \(2\cdot (3+6)\) (Complete this calculation in the two different ways present in the distributive law and verify that they are equal.)
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2.

What is the additive inverse of \(3\) in \(\ff_5\text{?}\) What is the multiplicative inverse of \(3\) in \(\ff_7\text{?}\) Explain your answers.
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3.

Theorem A.0.6 says that \(\zz_9\) is not a field because 9 is not prime. Identify a nonzero element of \(\zz_9\) that does not have a multiplicative inverse and explain why it does not have an inverse.
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Exercises Exercises

1.

Carry out the following calculations in \(\zz_8\text{.}\) (Remember that your answer for calculations in \(\zz_n\) should be a number between \(0\) and \(n-1\text{.}\))
  1. \(\displaystyle 6 + 7\)
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  2. \(\displaystyle 3 + 7 + 5\)
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  3. \(\displaystyle 7 \cdot 3\)
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  4. \(\displaystyle 6 \cdot 4 \cdot 3\)
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  5. \(\displaystyle (-3)\cdot (5 + 3 + (-4))\)
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2.

We know that since \(12\) is not a prime, \(\zz_{12}\) is not a field. In particular, the axiom about multiplicative inverses does not hold. For each nonzero element of \(\zz_{12}\text{,}\) determine whether or not it has a multiplicative inverse. If the element has a multiplicative inverse, state that inverse.
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3.

For each nonzero element of \(\ff_{11}\text{,}\) find the multiplicative inverse.
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Writing Exercises

5.
Consider Example 2.1.6. Which of the field axioms for \(\qq\) hold “by inheritance” and for which of the axioms is there something that needs to be proved? Put each of the field axioms into one of these two categories. For each axiom that doesn’t hold merely “by inheritance,” prove that it holds for \(\qq\text{.}\)
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6.
Let \(A\) be a set and let \(n\) be a nonnegative integer. Define a polynomial \(p(x)\) of degree \(n\) over \(A\) by
\begin{equation*} p(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n, \end{equation*}
where \(a_i \in A\) for all \(i\text{.}\) Define \(A[x]\) as the set of polynomials of any degree over \(A\text{.}\)
  1. Is \(\zz[x]\) a field? Justify your answer.
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  2. Is \(\rr[x]\) a field? Justify your answer.
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Answer.
  1. No, \(\zz[x]\) is not a field. All of the integers are contained in \(\zz[x]\text{,}\) and since (for example) \(2 \in \zz\) has no multiplicative inverse in \(\zz\text{,}\) it will not have a multiplicative inverse in \(\zz[x]\text{.}\)
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  2. No, \(\rr[x]\) is not a field. The element \(x \in \rr[x]\) has no multiplicative inverse. We can argue this by contradiction. If \(b = a_0 + a_1x + a_2x^2 + \cdots\) was the multiplicative inverse of \(x\text{,}\) then
    \begin{equation*} 1 = xb = a_0x + a_1x^2 + a_2x^3 + \cdots\text{.} \end{equation*}
    But since the constant term on the right side of this equation is \(0\) and the constant term on the left side is \(1\text{,}\) and since those are not equal, we have a contradiction.
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7.
In this problem we consider “adding” an element to \(\ff_3\text{.}\) If \(\ff_3[\alpha]\) is defined by
\begin{equation*} \ff_3[\alpha] = \{a + b\alpha \mid a,b \in \ff_3, \alpha^2 = 2\}, \end{equation*}
is \(\ff_3[\alpha]\) a field? Justify your answer.
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8.
In this problem we consider “adding” an element to \(\ff_5\text{.}\)
  1. If \(\ff_5[\alpha]\) is defined by
    \begin{equation*} \ff_5[\alpha] = \{a + b\alpha \mid a,b \in \ff_5, \alpha^2 = 4\}, \end{equation*}
    is \(\ff_5[\alpha]\) a field? Justify your answer.
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  2. If \(\ff_5[\alpha]\) is defined by
    \begin{equation*} \ff_5[\alpha] = \{a + b\alpha \mid a,b \in \ff_5, \alpha^2 = 2\}, \end{equation*}
    is \(\ff_5[\alpha]\) a field? Justify your answer.
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Answer.
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  2. Yes, this is a field. The most difficult axiom to check is the one about multiplicative inverses. It turns out that the multiplicative inverse of \(a + b\alpha\) is \(c + d\alpha\) where
    \begin{equation*} c = \frac{1}{a^2-2b^2}, \hspace{6pt} \text{and} \hspace{6pt} d = -\frac{b}{a^2-2b^2}\text{.} \end{equation*}
    One needs to check that for all non-zero elements \(a + b\alpha\) of \(\ff_5[\alpha]\) we have \(a^2-2b^2 \neq 0\text{.}\) But this is relatively easy to verify.
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9.
Define \(\qq[\sqrt{2}]\) by
\begin{equation*} \qq[\sqrt{2}] = \{a + b\sqrt{2} \mid a,b \in \qq\}. \end{equation*}
Is \(\qq[\sqrt{2}]\) a field? Justify your answer.
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10.
Consider the set \(\rr^2\) with operations defined as follows:
\begin{align*} (a,b) \oplus (c,d) \amp = (a+c,b+d)\\ (a,b) \otimes (c,d) \amp = (ac - 2bd, ad+bc)\text{.} \end{align*}
Is \(\rr^2\) with these operations a field? Justify your answer.
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Answer.
Yes, this is a field. Some of the axioms are quite tedious to check. It is perhaps worth noting that the multiplicative inverse of \((a,b)\) is
\begin{equation*} \left(\frac{a}{a^2+2b^2}, -\frac{b}{a^2+2b^2}\right)\text{.} \end{equation*}
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12.
Solution.
Let \(b \in \ff\) be a non-zero element with multiplicative inverse \(b'\text{.}\) Suppose that the element \(a \in \ff\) also has the properties of a multiplicative inverse of \(b\text{.}\) This means that \(ab=1\text{.}\) If we multiply both sides of this equation by \(b'\text{,}\) we have
\begin{align*} (ab)b' \amp =b'\\ a(bb') \amp =b'\\ a(1) \amp =b'\\ a \amp =b'\text{.} \end{align*}
This proves that \(a=b'\text{,}\) so multiplicative inverses are unique.
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