6 Algebraic structures
We complete our exposition of the mathematical background by discussing some common algebraic structures, which generalize what we have seen in the previous section. Informally, an algebraic structure is a set with one or more operations on the elements of the set. The properties of these operations determine the kind of structure that we have. In this section, we will learn about two of these:
Groups.
Finite fields.
6.1 Groups
Groups are the simplest algebraic structure that we will study, since they only have one operation.
Definition 6.1 A group is a pair \((\mathbb{G},\circ)\), where \(\mathbb{G}\) is a set and \(\circ\) is an operation on \(\mathbb{G}\), that is, a function \[\begin{aligned} \circ : \phantom{a} & \mathbb{G}\times \mathbb{G}& \rightarrow & \phantom{a}\mathbb{G}\\ & (x,y) & \mapsto & \phantom{a} x\circ y, \end{aligned}\] which additionally satisfies the following properties:
Associative law: \((x\circ y)\circ z = x\circ(y\circ z)\) for all \(x,y,z\in \mathbb{G}\).
Existence of identity: there exists \(e\in \mathbb{G}\) such that \(e\circ x=x\circ e=x\) for all \(x\in \mathbb{G}\). Such element \(e\) is called the identity element.
Existence of inverse: for any \(x\in \mathbb{G}\), there exists \(y\in \mathbb{G}\) such that \(x\circ y = y\circ x = e\), where \(e\) is the identity element. Such \(y\) is called the inverse of \(x\).
A group is said to be commutative (or abelian) if it satisfies the following additional property:
- Commutativity: \(x\circ y=y \circ x\) for all \(x,y\in \mathbb{G}\).
The order of a group is the number of elements in \(\mathbb{G}\), if the set is finite, and infinite otherwise, and is denoted by \(ord(\mathbb{G})\).
When there is no ambiguity about the operation, we will refer to the group \(\mathbb{G}\), instead of \((\mathbb{G},\circ)\), for simplicity. Most of the time, the group operation will be (possibly modular) addition or multiplication, although these are not the only possible cases. Let us consider some examples, and see whether they are groups or not.
\((\mathbb{Z}_{\geq0},+)\), where \(\mathbb{Z}_{\geq0}\) is the set of non-negative elements of \(\mathbb{Z}\), and \(+\) is integer addition. Integer addition is associative, and there is an identity element \(0\). However, there is no \(x\in\mathbb{Z}_{\geq0}\) such that \[1+x=0,\] and therefore \((\mathbb{Z}_{\geq0},+)\) is not a group.
\((\mathbb{Z},+)\) is a group of infinite order, since the operation is still associative, there is an identity element \(0\), and for any \(x\in\mathbb{Z}\), there exists \(y=-x\in\mathbb{Z}\) such that \[x+y=y+x=0.\] For similar reasons, the pair \((\mathbb{Z}_n,+)\), for \(n\in\mathbb{N}\), is also a group, of order \(n\).
For \(n\in\mathbb{N}\), the pair \((\mathbb{Z}_n^*,\cdot)\), where \(\cdot\) is multiplication modulo \(n\), is a group of order \(\varphi(n)\). The operation is clearly associative, \(1\) is an identity element for multiplication, and every element \(x\in\mathbb{Z}_n^*\) has an inverse \(x^{-1}\in\mathbb{Z}_n^*\), by definition of \(\mathbb{Z}_n^*\).
The pair \((\mathbb{Z}_2\times\mathbb{Z}_3,+)\), where \(+\) means component-wise addition in their respective moduli, is a group of order \(6\). Indeed, it is easy to see that a product of two groups is a group.
Moreover, since the operations are commutative, all of these groups are commutative.
Exercise 6.1 Explain why none of the following is a group:
\((\mathbb{Z},\cdot)\), where \(\cdot\) is integer multiplication.
\((\mathbb{Z}_n,\cdot)\), where \(\cdot\) is multiplication modulo \(n\), for \(n\in \mathbb{N}\).
Group notation. In the above examples, you can observe that the notation differs from case to case, depending on the nature of the operation. For example, given some element \(x\), we denote its inverse with respect to addition by \(-x\), and its inverse with respect to multiplication by \(x^{-1}\). Applying the operation to \(n\) copies of the same element \(x\) is represented by \(nx\) in additive notation, and by \(x^n\) in multiplicative notation. Even though not all group operations are addition or multiplication, it is common to adopt their notation for a generic group operation.21 In these notes, we will often use multiplicative notation for generic groups. That is, we denote the identity element by \(1\), the inverse of \(x\) by \(x^{-1}\), and the operation of \(x\) and \(y\) by \(x\cdot y\) or simply \(xy\).
Observe that, if a group \(\mathbb{G}\) contains an element \(g\), then the fact that \(g^2=gg\) implies that \(g^2\in\mathbb{G}\) too. Similarly, \(g^3=g^2g\in \mathbb{G}\), and so on. It is easy to generalize this idea, and conclude that, if \(g\in \mathbb{G}\), then \(g^n\in \mathbb{G}\) for all \(n\in\mathbb{N}\). That is, a single element \(g\) generates many elements of a group. This leads us to the concept of cyclic groups, which are those that contain only the powers of a single element.
Definition 6.2 A group \(\mathbb{G}\) is said to be cyclic if there exists \(g\in \mathbb{G}\) such that \[\mathbb{G}=\{g^n\mid n\in\mathbb{Z}_{\geq0}\}.\] This group is also denoted by \(\langle g \rangle\), and is called the group generated (or spanned) by \(g\), and \(g\) is called a generator of \(\mathbb{G}\). Note that, in a group with additive notation, we would write instead \[\mathbb{G}=\{nx\mid n\in\mathbb{Z}_{\geq0}\}.\]
The name of these groups come from their often cyclic nature. We illustrate this with the example of the group \(\mathbb{Z}_5\) with addition modulo \(5\). We claim that \(\mathbb{Z}_5\) is generated by \(2\). Indeed, \[\begin{aligned} & 2\cdot 0\bmod 5 = 0\\ & 2\cdot 1\bmod 5 = 2\\ & 2\cdot 2\bmod 5 = 4\\ & 2\cdot 3\bmod 5 = 6 \bmod 5 = 1\\ & 2\cdot 4\bmod 5 = 8 \bmod 5 = 3\\ & 2\cdot 5\bmod 5 = 10 \bmod 5 = 0\\ & 2\cdot 6\bmod 5 = 12 \bmod 5 = 2\\ & 2\cdot 7\bmod 5 = 14 \bmod 5 = 4\\ & \vdots \end{aligned}\]
We make two observations: the first is that all the elements of \(\mathbb{Z}_5\) are reached through powers of \(2\), thus we have proven that it is generated by \(2\). The second is that, after \(2\cdot 5\), it is clear that the numbers start to repeat in a fixed order. That is, the powers of \(2\) cycle through \(\mathbb{Z}_5\) in the order \(0,2,4,1,3\). Note that generators are not unique. For example, \(\mathbb{Z}_5\) could be generated by \(1\) as well, although not by \(0\).
Moreover, although we will not get into the technical details, we informally mention that any cyclic group of finite order \(n\) behaves like the additive group \(\mathbb{Z}_n\), and any cyclic group of infinite order behaves like \(\mathbb{Z}\).
Exercise 6.2 Decide whether the group \(\mathbb{Z}_2\times\mathbb{Z}_3\), with component-wise modular addition, is a cyclic group.
Groups might contain smaller groups inside, that are consistent with the same operation.
Definition 6.3 Let \(\mathbb{G}\) be a group. A subgroup \(\mathbb{H}\) of \(\mathbb{G}\) is a subset of \(\mathbb{G}\) that contains the identity element and such that \(\mathbb{H}\) forms a group with the operation of \(\mathbb{G}\) restricted to \(\mathbb{H}\). That is, for any \(x,y\in \mathbb{H}\), we have that \(x^{-1}\in\mathbb{H}\) and \(xy\in \mathbb{H}\).
For example, the set of even integers is a subgroup of \((\mathbb{Z},+)\), since the addition of even numbers is even. It is easy to see that subgroups are groups too.
Exercise 6.3 Decide which of these subsets are subgroups of \((\mathbb{Z}_4,+)\): \[\{0\},\quad\qquad\{0,2\},\quad\qquad\{1,3\},\quad\qquad\{0,1,3\}.\]
It is easy to see that, for any \(x\in\mathbb{G}\), the group \(\langle x\rangle\) is a subgroup of \(\mathbb{G}\). Note that this might not be the whole \(\mathbb{G}\). This allows us to define the order of an element as follows.
Definition 6.4 Let \(\mathbb{G}\) be a group, and \(x\in\mathbb{G}\). We define the order of \(x\) as the order of \(\langle x\rangle\), and denote it by \(ord(x)\).
The following Sage code defines the additive group \((\mathbb{Z}_9,+)\), computes a generator, and then computes the order of each element, also explicitly providing the cycle that corresponds to that element.
G = AdditiveAbelianGroup([9])
g = G_add.gens()[0] # The [0] is necessary because, technically,
# .gens() returns a list.
print(str(G)+", generator: "+str(g))
for x in G:
print("x = "+str(x)+", ord(x) = "+str(x.order()))
for i in range(x.order()+1):
print(i*x)If \(\mathbb{H}\) is a subgroup of \(\mathbb{G}\), it is clear that the order of \(\mathbb{H}\) will be smaller than the order of \(\mathbb{G}\), since the former is contained in the latter. But even more, it will necessarily be a divisor, which is something you may have observed after running the above code.
Proposition 6.2 (Lagrange’s theorem) Let \(\mathbb{G}\) be a finite group.
For any subgroup \(\mathbb{H}\) of \(\mathbb{G}\), \(ord(\mathbb{H})\mid ord(\mathbb{G})\).
For any \(x\in\mathbb{G}\), \(ord(x)\mid ord(\mathbb{G})\).The order of a group plays an important role in the group operation.
Exercise 6.4 Verify that Euler’s theorem holds for the group \(\mathbb{Z}_7^*\), with the operation multiplication modulo \(7\).
The exercise can also be solved with Sage, using the following code.
G = Integers(7).unit_group() # Multiplicative group Z_7^*.
for x in G:
print(x^(G.order()))These propositions give us an easy way to check whether an element \(x\in\mathbb{G}\) is a generator. Recall that the order of \(x\) tells us how many elements there are in \[\langle x\rangle = \{x^n\mid n\in\mathbb{Z}_{\geq0}\},\] that is, the number of steps in the cycle of powers of \(x\) before going back to \(1\) and repeating elements. If it were the case that \(ord(x)=ord(\mathbb{G})\), this would mean that a the cycle is as big as \(\mathbb{G}\), and therefore it must be that \(\langle x\rangle = \mathbb{G}\). On the other hand, if \(ord(x)<ord(\mathbb{G})\), this would mean that not all elements of \(\mathbb{G}\) can be obtained as powers of \(x\), and thus \(x\) would not generate \(\mathbb{G}\).
Hence, we can check whether \(x\) generates \(\mathbb{G}\) by computing \(ord(x)\) and comparing it with \(ord(\mathbb{G})\). To do so, a naive approach would be to iteratively compute \[x,x^2,x^3,x^4,x^5,\dots\] until some \(n\) is found such that \[x^n=1,\] and we would have that \(n=ord(x)\). However, note that this approach takes time linear in \(ord(\mathbb{G})\). When \(ord(\mathbb{G})\) is easy to factor, a more sensible approach is to use Lagrange’s theorem, which tells us that \[ord(x)\mid ord(\mathbb{G}).\] This allows us to drastically reduce the candidates to the powers \[x^{\frac{ord(\mathbb{G})}{d}},\] for all \(d\mid ord(\mathbb{G})\). The smallest exponent that produces a \(1\) will be the order of \(x\). In particular, if \(ord(G)\) is prime, then there are only two candidates: \(x\) and \(x^{ord(\mathbb{G})}\). Hence in this case, if \(x\neq1\), necessarily \(x\) is a generator. That is, we have proven the following result.
We conclude the section on groups by looking at some specific properties of the groups \(\mathbb{Z}_n\) and \(\mathbb{Z}_n^*\).
Moreover, a direct consequence of Euler’s theorem and the proposition above is known as Fermat’s little theorem.22
6.2 Finite fields
A field is a more complex algebraic structure, since it is equipped with two operations, both of which work similarly to a group operation.23
Definition 6.5 A field is a triple \((\mathbb{F},\circ,\ast)\), where \(\mathbb{F}\) is a set and \(\circ\) and \(\ast\) are operations on \(\mathbb{F}\), that is, functions \[\begin{aligned} \circ : \phantom{a} & \mathbb{F}\times \mathbb{F}& \rightarrow & \phantom{a}\mathbb{F}& \qquad\qquad & \ast :& \mathbb{F}\times\mathbb{F}& \rightarrow & \mathbb{F}\\ & (x,y) & \mapsto & \phantom{a} x\circ y, & & & (x,y) & \mapsto & x\ast y, \end{aligned}\] which additionally satisfy the following properties:
\((\mathbb{F},\circ)\) is a commutative group, with identity element denoted by \(0\).
\((\mathbb{F}\setminus\{0\},\ast)\) is a commutative group, where \(\mathbb{F}\setminus\{0\}\) is the set \(\mathbb{F}\) except for the identity element of \(\circ\).
Distributive law: \(x\ast(y\circ z)=(x\ast y)\circ(x\ast z)\) for all \(x,y,z\in\mathbb{F}.\)
Since we have two operations at the same time, we will adopt the additive and multiplicative notation from groups to represent each of them, respectively. That is, we will think of the first operation of a field as a form of addition and the second as a form of multiplication. On input \(x,y\), we write the result of the first operation by \(x+y\), and the result of the second by \(xy\). The identity elements of each operation are denoted by \(0\) and \(1\), respectively. The inverse of \(x\) with respect to the first operation is denoted by \(-x\). The inverse of \(x\) with respect to the second operation is denoted by \(x^{-1}\). The following table summarizes the notation:
| Operation | Operation on input \(x,y\) | Identity element | Inverse of \(x\) |
|---|---|---|---|
| \(+\) | \(x+y\) | \(0\) | \(-x\) |
| \(\cdot\) | \(xy\) | \(1\) | \(x^{-1}\) |
We consider some examples:
\(\mathbb{Q}\), \(\mathbb{R}\) and \(\mathbb{C}\), with usual addition and multiplication, are fields. The properties are easy to check. Identity elements are \(0\) and \(1\). The additive inverse of \(x\) is \(-x\), and the multiplicative inverse of \(x\) is \(1/x\).
Consider \(\mathbb{Z}_9\), with addition and multiplication modulo \(9\). It satisfies most of the properties of a field, but we will see that some elements have no inverse with respect to the second operation. The following table gives the multiplicative inverses of each element:
\(x\) \(0\) \(1\) \(2\) \(3\) \(4\) \(5\) \(6\) \(7\) \(8\) \(x^{-1}\) \(-\) \(1\) \(5\) \(-\) \(7\) \(2\) \(-\) \(4\) \(8\) Note that some elements are not invertible, in this case \(0\), \(3\) and \(6\). Therefore, \(\mathbb{Z}_9\) is not a field.
In Sage, the following code can be used to check whether an element of \(\mathbb{Z}_n\) has a multiplicative inverse.
R = Integers(n) # Z_n with addition and
# multiplication modulo n.
for x in R:
print(x,x.is_unit())In this course we will be interested in finite fields,24 so what are some examples of those? Could it be, for example, that \(\mathbb{Z}_n\) is a field for some \(n\in\mathbb{N}\)? We already know that this will not be true for all \(n\in\mathbb{N}\), since we saw that some elements of \(\mathbb{Z}_9\) do not have an inverse.
The only finite fields that exist have a number of elements that is either a prime number or a power of a prime number.
Moreover, given a prime \(p\) and \(k\in\mathbb{N}\), there is essentially only one field with \(p^k\) elements, although this same field can have different representations.
One might think that these extension fields \(\mathbb{F}_{p^k}\) correspond to \(\mathbb{Z}_{p^k}\), but actually this is not true. Again, we recall the example in which we showed that \(\mathbb{Z}_9=\mathbb{Z}_{3^2}\) is not a field, because some elements are not invertible. So, if \(\mathbb{F}_{p^k}\) is not \(\mathbb{Z}_{p^k}\), what is it?
It turns out that we need some more involved tools to describe these. An element of the field \(\mathbb{F}_{p^k}\) is not represented by an integer. Instead, we use a polynomial \[r_{k-1}x^{k-1}+\dots +r_1x+r_0,\] for some coefficients \(r_{k-1},\dots,a_0\in\mathbb{F}_p\). Note that the degree of the polynomial is \(k-1\) for representing an element of \(\mathbb{F}_{p^k}\). Since there are \(k\) coefficients, and each of these has \(p\) possible values, we see that there are in total \(p^k\) possible polynomials, and thus so far it makes sense to try to associate these with the \(p^k\) elements of \(\mathbb{F}_{p^k}\).
But we still need to specify how the operations work. For addition, we use usual polynomial addition, that is, \[\begin{split} R(X)+S(X)& =(r_{k-1}X^{k-1}+\dots +r_1X+r_0)+(s_{k-1}X^{k-1}+\dots +s_1X+s_0) = \\ & =(r_{k-1}+s_{k-1})X^{k-1}+\dots +(r_1+s_1)X+(r_0+s_0), \end{split}\] which is also a polynomial of degree at most \(k-1\), and it is easy to see that it verifies the conditions of a group operation. Note that the coefficients are in \(\mathbb{F}_p\), so the additions of the coefficients are reduced modulo \(p\) if necessary.
For multiplication, we consider multiplying the two polynomials. However, note that, in general, multiplying two polynomials \(R(X)\) and \(S(X)\) of degree \(k-1\) produces a polynomial \(R(X)S(X)\) of degree \(2k-2\), which is not in \(\mathbb{F}_{p^k}\) anymore. Our trick then is analogous to what we did in \(\mathbb{F}_p=\mathbb{Z}_p\). In that case, we reduced the result modulo a prime number \(p\). Now, we will reduce the result modulo a certain polynomial \(T(X)\) of degree \(k\). This means that we compute the polynomial division of \(R(X)S(X)\) by \(T(X)\), and keep the remainder, so that the result has degree at most \(k-1\).25 Since this is analogous to the case of integers, we extend the notation \[A(X)\bmod{T(X)}\] to denote the remainder of the polynomial division \(A(X)\) by \(T(X)\), and the notation \[A(X)\equiv B(X)\pmod{T(X)}\] to denote that \[(A(X)-B(X))\bmod{T(X)}=0,\] that is, \(A(X)-B(X)\) is a multiple of \(T(X)\).
But which polynomial \(T(X)\) should we use? Not every polynomial will yield a multiplication law of a field. We require an irreducible polynomial, that is, a polynomial that cannot be written as a product of two non-trivial polynomials with coefficients in \(\mathbb{F}_p\).
At first, this might seem like an obscure condition, but if you think about it, it is actually analogous to what happened for prime fields. In that case, we started from \(\mathbb{Z}_n\), and concluded that it is only a field when \(n\) is a prime, which is precisely a number that cannot be factored in a non-trivial way. In extension fields, we are representing elements by polynomials instead of integers, and so we look for the analogue of a prime number in this sense of lack of factorization. This leads us to irreducible polynomials, although a formal proof of this statement is beyond the scope of this course.
The choice of irreducible polynomial \(T(X)\) determines the multiplication law, but any of the outcomes can be seen as different representations of the same finite field of size \(p^k\).
Finally, inversion of elements in \(\mathbb{F}_{p^k}\) can be handled with the extended Euclidean algorithm por polynomials, similarly to the way it was used to invert integers modulo \(n\). We describe this algorithm below.
Definition 6.6 Given a polynomial \[a_kX^k+a_{k-1}X^{k-1}+\dots+a_1X+a_0,\] we define its leading coefficient as the coefficient of the highest-degree monomial, in this case \(a_k\). A polynomial with leading coefficient \(1\) is called monic.
Definition 6.7 Given two polynomials \(A(X)\) and \(B(X)\), we define their greatest common divisor as the highest-degree monic polynomial that divides both.
The Euclidean algorithm for polynomials is very similar to the analogous algorithm for integers, with an additional step to ensure that the output is a monic polynomial.
Compute the polynomial division of \(A(X)\) by \(B(X)\), obtaining \(Q(X),R(X)\) such that the \(0\leq \deg R<\deg B\) and \[A(X)=B(X)Q(X)+R(X).\]
If \(R(X)=0\), go to step 3. Otherwise, return to the previous step, replacing \(A(X)\) by \(B(X)\) and \(B(X)\) by \(R(X)\).
Multiply \(B(X)\) by the inverse of its leading coefficient, and output the result.
This will produce \(\gcd(A(X),B(X))\), from which we can go back through the procedure to find the expression in the proposition above. We show an example in \(\mathbb{F}_{5^4}\), with irreducible polynomial \(T(X)=X^4+4X^2+4X+2\). \[A(X)=X^3+3X^2+4, \qquad\qquad B(X)=X^2+2X+2.\] Note that coefficients are in \(\mathbb{F}_5\), and thus all the operations performed on coefficients must be carried out modulo \(5\). We start by dividing \(A(X)\) by \(B(X)\), obtaining \[X^3+3X^2+4=(X^2+2X+2)(X+1)+(X+2)\] Since the remainder is not \(0\), we compute the next polynomial division, obtaining \[X^2+2X+2=(X+2)X+2.\] Finally, \[(X+2)=2(3X+1)+0.\] Since the last remainder is \(0\), we conclude that the greatest common divisor of \(A(X)\) and \(B(X)\) is \(2\) multiplied by the inverse of its leading coefficient, so in this case \[\gcd(A(X),B(X))=2\cdot2^{-1}=1.\] Again, by undoing these steps, we can find the expression in Proposition 6.10: \[2=(X^2+2X+2)-(X+2)X.\] We also have that \[X+2=(X^3+3X^2+4)-(X^2+2X+2)(X+1),\] and thus, combining the two expressions, \[\begin{aligned} 2 & = (X^2+2X+2)-\left((X^3+3X^2+4)-(X^2+2X+2)(X-1)\right)X = \\ & = (X^2+2X+2)-(X^3+3X^2+4)X + (X^2+2X+2)(X^2+X) = \\ & = (-X)(X^3+3X^2+4)+(X^2+X+1)(X^2+2X+2). \end{aligned}\] Multiplying both sides by \(2^{-1}\bmod{5}=3\), we have that \[\begin{aligned} 1 & \equiv (-3X)(X^3+3X^2+4)+(3X^2+3X+3)(X^2+2X+2) \equiv \\ & \equiv (2X)(X^2+3X^2+4)+(3X^2+3X+3)(X^2+2X+2). \end{aligned}\] Therefore, \[U(X)=2X,\qquad\qquad V(X)=3X^2+3X+3.\]
As in the case of integers, we can compute the inverse of \(A(X)\) in \(\mathbb{F}_{p^k}\) by running the Extended euclidean algorithm on \(A(X)\) and the irreducible polynomial \(T(X)\), obtaining some polynomials \(U(X),V(X)\) such that \[1=A(X)U(X)+T(X)V(X),\] thus \[1\equiv A(X)U(X)\pmod{T(X)},\] and we find that \(U(X)\) is the inverse of \(A(X)\) in \(\mathbb{F}_{p^k}\).
We show to define and manipulate finite fields in Sage.
# Finite field of size 3^5. The optional argument modulus=p
# allows to specify the irreducible polynomial p to be used.
F = GF(3^5, "X")
F
# Get the polynomial p(X)=X into the field.
X = F("X")
# Show the irreducible polynomial being used.
F.modulus()
# Define two elements of F and operate with them.
a = F(X^4+X^2+2*X)
b = F(2*X^2+2*X^3+1)
# Addition
print("a + b = "+str(a+b))
# Multiplication
print("a * b = "+str(a*b))
# Inversion
print("a^{-1} = "+str(a^(-1)))A prime example of a different group operation, which appears very often in cryptography, is the group law of elliptic curves.↩︎
Not to be confused with Fermat’s last theorem, which states that there are no positive integer solutions for the equation \(x^n+y^n=z^n\) for any integer \(n>2\). This theorem has become famous due to its history: it was originally claimed by Pierre de Fermat in the 17th century. He wrote it in the margin of a book, claiming to know a proof but lacking the space to write it down. Such proof was never found, and the first known proof of the theorem appeared more than three centuries later, in the 1990s, when Andrew Wiles finally solved the problem, after the contributions of a long lineage of mathematicians.↩︎
Appendix A contains a more detailed introduction to this section. It first defines another algebraic structure called a ring, which is somewhat between a group and a field, and then defines fields as a particular type of ring. This appendix is not part of the content for the exam, and the main body of these notes is designed to be self-contained, but nevertheless we advise you to read Appendix A before reading about fields, since it provides a more natural introduction to the topic.↩︎
Sometimes also called Galois fields, after the mathematician Évariste Galois, who notably proved that there is no algebraic formula for solving polynomial equations of degree higher than \(4\), and then died after a duel at the age of 20.↩︎
See Appendix D.4 for a refresher on how to perform polynomial division.↩︎