## Section8.2A Review of Rings and Fields

The goal of this chapter is to develop a theory of fields that will help us answer questions like those about constructible numbers from the last section, as well as understand what fields can tell us about polynomials.

Before we can dive into the new details of this study, let's review what we know about rings (since fields are a type of ring) and the quotient structures that rings have. The next two activities ask you to go back over some topics already covered in the book, and you should feel free to review your notes or the earlier sections to help work through them.

### Worksheet8.2.1Activity: Review of Rings and Quotient Rings

The goal of this activity is to remind ourselves of basic but crucially important definitions we will need in our study of fields.

You are asked to provide definitions. Some definitions will include terms that also should be defined. Make sure that you know what every word in a definition means (if not, provide definitions for those words). For example, a field is a commutative division ring. If you have not yet defined commutative ring and division ring, you should say what these mean.

###### 1.

Give a definition of a ring.

###### 2.

What is a commutative ring with unity? How is this different from a ring? (Note, “unity” is also sometimes called “identity”.)

###### 3.

What is a commutative division ring? What does “division” refer to here, and how is this different from a ring in general?

###### 4.

Give the definition of an integral domain. How does this relate to the other types of structures you defined above?

###### 5.

What is an ideal? What is the difference between an ideal and a subring?

###### 6.

Consider the integers $\Z$ (an integral doamin, right?). What does the notation $\langle 3 \rangle$ mean? What sort of thing is this? What is $\langle r \rangle$ in general?

###### 7.

What is $\Q[x]\text{?}$ Then give an example of an ideal in $\Q[x]\text{,}$ using proper notation and by listing out some of the elements in the ideal.

###### 8.

Give the definition of a quotient ring (i.e. a factor ring). What do elements of a quotient ring look like? How are the operations defined?

###### 9.

Illustrate you you wrote about quotient rings above using two examples: First, $\Z/\langle 3 \rangle\text{,}$ and then $\Q[x]/\langle x^2 + 1\rangle\text{.}$ How many elements are in each of these quotient rings? What do the elements look like? Show how to add/multiply elements.

### Worksheet8.2.2Activity: Review of the Euclidean Algorithm

The goal of this activity is to remember how to use the Eulcidean Algorithm to find the greatest common divisor of two elements in a ring (numbers or polynomials, for us) and write the gcd as a linear combination of the two elements (which Bezout's lemma tells us we can do).

###### Example8.8.

Let's find the gcd of $945$ and $2415\text{.}$ Repeatedly use the division algorithm:

\begin{align*} 2415 & = 945 \cdot 2 + 525\\ 945 & = 525 \cdot 1 + 420\\ 525 & = 420 \cdot 1 + 105\\ 420 & = 105 \cdot 4 + 0\text{.} \end{align*}

Check: $105$ divides all the quotients and remainders, and any other divisor of $945$ and $2415$ would also divide $105\text{.}$ Therefore, $\gcd( 945, 2415 ) = 105\text{.}$

Now work backwards to obtain numbers $r$ and $s$ such that $945 r + 2415 s = 105\text{.}$

\begin{align*} 105 & = 525 + (-1) \cdot 420\\ & = 525 + (-1) \cdot [945 + (-1) \cdot 525]\\ & = 2 \cdot 525 + (-1) \cdot 945\\ & = 2 \cdot [2415 + (-2) \cdot 945] + (-1) \cdot 945\\ & = 2 \cdot 2415 + (-5) \cdot 945\text{.} \end{align*}

So $r = -5$ and $s= 2\text{.}$

###### 1.

Find the greatest common divisor of 471 and 564 using the Euclidean Algorithm and then find integers $r$ and $s$ such that $\gcd(471,564) = 471r+564s\text{.}$

###### 2.

In the quotient ring $\Z/\langle 564 \rangle\text{,}$ find an element $a + \langle 564\rangle$ such that $(a+\langle 564\rangle)(471 + \langle 564\rangle) = 3 + \langle 564 \rangle\text{.}$ Explain why the previous question is helpful here.

###### 3.

Is $471 + \langle 564\rangle$ a unit in $\Z/\langle 564\rangle\text{?}$ Explain.

###### 4.

In $\Q[x]\text{,}$ find the gcd of the polynomials $a(x) = x^3 + 1$ and $b(x) = x^4 + x^3 + 2x^2 + x - 1\text{.}$ Then express the gcd as a combination of the two polynomials (as in Bezout's lemma).

###### 5.

Find the greatest common divisor of $x^{24}-1$ and $x^{15}-1$ in $\Q[x]\text{,}$ and then express the gcd as a combination of the two polynomials.

###### 6.

Find a coset $a(x) + \langle x^{24}-1\rangle$ of $\Q[x]/\langle x^{24}-1\rangle$ such that $(a(x) + \langle x^{24}-1\rangle)(x^{15}-1 + \langle x^{24}-1\rangle) = x^3-1 + \langle x^{24}-1\rangle\text{.}$

### Subsection8.2.3Fields for Polynomials

Recall that a field is a commutative division ring. That is, a field is a type of ring (i.e., a set together with two operations, addition and multiplication, satisfying the ring axioms). The ring is commutative (meaning multiplication commutes; addition is commutative in all rings), and every non-zero element in the ring has a multiplicative inverse.

A few examples of fields so far include the rational numbers $\Q\text{,}$ as well as the real numbers $\R$ and the complex numbers $\C\text{.}$ We have also considered finite fields such as $\Z_3\text{,}$ $\Z_5\text{,}$ and $\Z_7$ (but not $\Z_6$ which is not even an integral domain).

In fact, we have seen other fields as well: the quotient ring $\Q[x]/\langle x - 1\rangle$ is a field, which happens to be isomorphic to $\Q\text{.}$ Another example is $\Q[x]/\langle x^2 - 2\rangle\text{,}$ which is not isomorphic to $\Q\text{,}$ but still happens to be a field. When we considered quotient rings like these before, we focused on their structure as a quotient ring, not worrying too much about its structure as a field. That is about to change.

A reasonable place to start is to consider how different fields relate to each other. This is not unlike studying subgroups to better understand groups, or subrings and ideals to better understand rings. One notable difference is that with fields, we often think about expanding a smaller field into a larger one, rather than restricting a larger group into a (smaller) subgroup.

Given some field $F\text{,}$ we can ask whether it is contained in a larger field. Think of the rational numbers, which reside inside the real numbers, while in turn, the real numbers live inside the complex numbers. We would say that the rational numbers are a subfield of the real numbers, or say that the real numbers are a extension field of the rational numbers. Indeed, we often think of the real numbers as the field you get by adding all the “gaps” in the number line to the rational numbers; we extend the rationals to get the reals. Similarly, if you start with the real numbers and “throw in” the imaginary number $i\text{,}$ you get the complex numbers (an extension field of both $\Q$ and $\R$). The smaller field $F$ is called the base field. We write $F \subset E\text{.}$

Are there other fields hanging out around these traditional choices? Is there a field between $\Q$ and $\R\text{?}$ We will see that the answer to this is an overwhelming yes! Most of what we mean by field theory is precisely the study of fields that are extensions of $\Q\text{.}$

It is not at all obvious where to start looking for fields between $\Q$ and $\R\text{.}$ We could start with $\R$ and eliminate numbers, but this does not do much for us. We really want to extend $\Q\text{.}$ What can we add to this already infinite set? We choose to add roots of polynomials! This will have the added benefit of helping us to study the polynomials as well.

More specifically if we are given a field $F$ and a polynomial $p(x) \in F[x]\text{,}$ we can ask whether or not we can find a field $E$ containing $F$ such that $p(x)$ factors into linear factors over $E[x]\text{.}$ For example, if we consider the polynomial

\begin{equation*} p(x) = x^4 -5 x^2 + 6 \end{equation*}

in $\Q[x]\text{,}$ then $p(x)$ factors as $(x^2 - 2)(x^2 - 3)\text{.}$ However, both of these factors are irreducible in $\Q[x]\text{.}$ If we wish to find a zero of $p(x)\text{,}$ we must (and can) go to a larger field. Certainly the field of real numbers will work, since

\begin{equation*} p(x) = (x - \sqrt{2} ) (x + \sqrt{2} )( x - \sqrt{3})(x + \sqrt{3})\text{.} \end{equation*}

It is possible to find a smaller field in which $p(x)$ has a zero, namely

\begin{equation*} {\mathbb Q }( \sqrt{2} ) = \{ a + b \sqrt{2} : a, b \in {\mathbb Q} \}\text{.} \end{equation*}

Our goal is to compute and study such fields for arbitrary polynomials over a field $F\text{.}$

Before proceeding though, we should probably convince ourselves that $E = \Q(\sqrt{2})$ really is a field. First, why is this a ring? Well, $E$ is a subset of $\R\text{,}$ so all we need to check is that it is a subring of $\R\text{.}$ If we take two elements $a+b\sqrt{2}$ and $c + d \sqrt{2}\text{,}$ is their sum, difference, and product still in $E\text{?}$ The only one of these that is not obvious is the product. But,

\begin{equation*} (a+b\sqrt{2})(c+d\sqrt{2}) = ac+2bd + (ad + bc)\sqrt{2}\text{,} \end{equation*}

which has the correct form to be in $E\text{.}$

What about multiplicative inverses though? Does every $a+b\sqrt{2}\text{,}$ with $a$ and $b$ not both zero, have an inverse? In this example, it is not too difficult to see the answer is yes. In fact, we can take

\begin{equation*} (a+b\sqrt 2)\inv = \frac{a}{a^2 - 2b^2} - \frac{b}{a^2 - 2b^2}\sqrt{2}\text{.} \end{equation*}

This was found by rationalizing the denominator of $\frac{1}{a+b\sqrt{2}}$ by multiplying by $a-b\sqrt{2}$ in both the numerator and denominator. That works fine for this simple example, but we will need to think about other methods for more complicated extensions.

Here is another example:

###### Example8.9.

Let's start with $\Q(\sqrt{2})$ as our base field, and take another extension. That is, let

\begin{equation*} F = {\mathbb Q}( \sqrt{2}\,) = \{ a + b \sqrt{2} : a, b \in {\mathbb Q} \} \end{equation*}

and let $E = {\mathbb Q }( \sqrt{2} + \sqrt{3}\,)$ be the smallest field containing both ${\mathbb Q}$ and the real number $\sqrt{2} + \sqrt{3}\text{.}$ For now, assume both $E$ and $F$ are fields, so they are both extension fields of the rational numbers. We claim that $E$ is an extension field of $F\text{.}$ To see this, we need only show that $\sqrt{2}$ is in $E\text{.}$ Since $\sqrt{2} + \sqrt{3}$ is in $E\text{,}$ $1 / (\sqrt{2} + \sqrt{3}\,) = \sqrt{3} - \sqrt{2}$ must also be in $E\text{.}$ Taking linear combinations of $\sqrt{2} + \sqrt{3}$ and $\sqrt{3} - \sqrt{2}\text{,}$ we find that $\sqrt{2}$ and $\sqrt{3}$ must both be in $E\text{.}$

Note that this also shows that $E$ is an extension field of $\Q(\sqrt{3})\text{.}$ Later we will consider how $E$ relates to the fields $\Q(\sqrt{2}, \sqrt{3})\text{,}$ the smallest field that contains both $\sqrt{2}$ and $\sqrt{3}\text{.}$

The example above make sense using real numbers we already know about. Our next goal will be to simply add a arbitrary root of a polynomial, perhaps without even knowing which real (or complex) number that root is. Surprisingly, this will work! To get a feel for this, attempt the problems on the following activity.

### Worksheet8.2.4Activity: Extending Fields to Factor

Goal: Build the smallest field possible in which $p(x) = x^3 + 3x^2 - x + 2$ is NOT irreducible.

Note that $p(x)$ is irreducible over $\Q$ because it has no roots in $\Q$ (why is this and why is that enough?). So let's invent a new number, call it $\s\text{,}$ and insist that $\s$ is a root of $p(x)\text{.}$ Then consider the smallest field $E$ larger than $\Q$ that also contains $\s\text{.}$

###### 1.

List five elements in $E$ that are NOT already in $\Q\text{.}$

Solution

For example, $\s$ (which is not in $\Q$ because, by the rational roots theorem, there are no roots to $p(x)$ in $\Q$), as well as $\s+1\text{,}$ $4\s\text{,}$ $\s^2\text{,}$ and $3+\frac{1}{2}\s - \s^2\text{.}$

###### 2.

The element $\s^3$ is in $E\text{,}$ but this can also be written using smaller powers of $\s\text{.}$ How?

Solution

Since $\s^3 + 3\s^2 - \s + 2 = 0$ we see that $\s^3 = -3\s^2 + \s - 2\text{.}$

###### 3.

Describe $E$ as a set using set builder notation. In other words, $E$ is the set of all elements of the form …

Solution

$E = \{a + b \s + c \s^2 \st a, b, c, \in \Q\}\text{.}$ Certainly everything of this form is in $E$ since $E$ must be closed under addition and multiplication. That nothing else is is a consequence of the previous question: since $\s^3$ can already be expressed as a linear combination of $1\text{,}$ $\s\text{,}$ and $\s^2\text{,}$ any power of $\s$ greater or equal to 3 can be as well.

###### 4.

Wait: why are we doing this? Our goal is for $p(x)$ to factor. Does it? What would one of the factors be?

Solution

Yes, we get $p(x) = (x-\s)(x^2 + (3+\s)x - 1 + 3\s + \s^2)\text{.}$ Do this by long division. The remainder might not look like zero right away, except that it is precisely $p(\s)\text{.}$ This is not a surprise: the division algorithm says that $p(x) = b(x)(x-\s) + r(x)$ where $r(x)$ is either the zero polynomial or has degree 0 (less than the degree of $(x-\s)$). Now plug in $\s$ for $x\text{.}$ Long division is just there to help us find $b(x)\text{.}$

###### 5.

Wait again: we want $E$ to be a field. Is it? What would we need to check?

Solution

This is a hard question that we will return to. Note that all we must check still is that every non-zero element in $E$ has a multiplicative inverse. That is, given $a+b\s + c\s^2\text{,}$ find an element $a' + b'\s + c'\s^2$ such that

\begin{equation*} (a+b\s+c \s^2)(a' + b'\s + c'\s^2) = 1\text{.} \end{equation*}

You could do this algebraically, although it is not an easy computation, and requires you to know that the polynomial $p(x)$ is irreducible. We will see why later.

###### 6.

List five elements in the quotient ring $\Q[x]/\langle p(x)\rangle$ (using the same $p(x)$ from the previous page). Remember, these will all be cosets.

Solution

For example, $x + \langle p(x) \rangle\text{,}$ $x+1 +\langle p(x) \rangle\text{,}$ $4x + \langle p(x) \rangle\text{,}$ $x^2 + \langle p(x) \rangle\text{,}$ and $3+\frac{1}{2}x - x^2 + \langle p(x) \rangle\text{.}$

###### 7.

The element $x^3+\langle p(x) \rangle$ is an element of $\Q[x]/\langle p(x) \rangle\text{,}$ but it can also be written as a “simpler” coset. How?

Solution

Since $x^3 + 3x^2 - x + 2 \in \langle p(x) \rangle$ we see that $x^3 + \langle p(x) \rangle = -3x^2 + x - 2 + \langle p(x) \rangle\text{.}$ You can get here either by dividing $x^3$ by $p(x)$ and looking for the remainder, or by asking, what would I need to add to $p(x)$ to get $x^3\text{.}$ That is because the coset $x^3 + \langle p(x) \rangle$ contains all the polynomials in $\Q[x]$ that result from adding $x^3$ to a multiple of $p(x)\text{.}$ One of these elements is $x^3$ (since the polynomial $0$ is a multiple of $p(x)$). So if we can find another polynomial $a(x)$ such that $a(x)$ added to a multiple of $p(x)$ gives $x^3\text{,}$ then $a(x) + \langle p(x)\rangle$ will be exactly the same coset as $x^3 + \langle p(x) \rangle\text{.}$

###### 8.

Describe $\Q[x]/\langle p(x) \rangle$ as a set using set builder notation. In other words, this quotient ring is the set of all cosets of the form …

Solution

$\Q[x]/\langle p(x) \rangle = \{a + bx + cx^2 + \langle p(x) \rangle \st a, b, c\in \Q\}\text{.}$ While we can certainly add a polynomial of degree greater to 2 to get a coset, we can always divide it by $p(x)$ and get a remainder of degree 2 or less.

This looks very much like the definition of $E\text{,}$ and in fact it suggests that $E \cong \Q[x]/\langle p(x) \rangle\text{.}$ This is correct, as we could prove using the fundamental homomorphism theorem. To do that, we would need to identify a surjective homomorphism from $\Q[x]$ onto $E$ for which $p(x)$ was the kernel.

###### 9.

Wait: if we want to show that $E$ is a field, and $E$ is basically the same as $\Q[x]/\langle p(x) \rangle\text{,}$ then we could just show $\Q[x]/\langle p(x) \rangle$ is a field. What would this mean? What do we need to verify?

Solution

We need to verify that for any coset $a+bx + cx^2 + \langle p(x) \rangle\text{,}$ there is a coset $a' + b'x + c'x^2 + \langle p(x) \rangle$ such that the product of the two cosets is the multiplicative identity coset, namely $1 + \langle p(x) \rangle\text{.}$ We will see how to do this next.