1 Fields and complex numbers
1.1 Fields
A field \(\mathbb{K}\) is roughly speaking a number system in which we can add, subtract, multiply and divide numbers, so that the expected properties hold. We will only briefly state the basic facts about fields. For a more detailed account, we refer to the algebra module.
Definition 1.1 • Field
A field consists of a set \(\mathbb{K}\) containing distinguished elements \(0_{\mathbb{K}}\neq 1_{\mathbb{K}},\) as well as two binary operations, addition \(+_{\mathbb{K}} : \mathbb{K}\times \mathbb{K}\to \mathbb{K}\) and multiplication \(\cdot_{\mathbb{K}} : \mathbb{K}\times \mathbb{K}\to \mathbb{K},\) so that the following properties hold:
Commutativity of addition \[x+_{\mathbb{K}}y=y+_{\mathbb{K}}x \quad \text{for all }x,y \in \mathbb{K}.\]
Commutativity of multiplication \[\tag{1.1} x\cdot_{\mathbb{K}}y=y\cdot_{\mathbb{K}}x \quad \text{for all }x,y \in \mathbb{K}.\]
Associativity of addition \[\tag{1.2} (x+_{\mathbb{K}}y)+_{\mathbb{K}} z=x+_{\mathbb{K}}(y+_{\mathbb{K}}z) \quad \text{for all }x,y,z \in \mathbb{K}.\]
Associativity of multiplication \[\tag{1.3} (x\cdot_{\mathbb{K}}y)\cdot_{\mathbb{K}} z=x\cdot_{\mathbb{K}}(y\cdot_{\mathbb{K}}z) \quad \text{for all }x,y,z \in \mathbb{K}.\]
\(0_\mathbb{K}\) is the identity element of addition \[\tag{1.4} x+_{\mathbb{K}}0_{\mathbb{K}}=0_{\mathbb{K}}+_{\mathbb{K}}x=x \quad \text{for all }x \in \mathbb{K}.\]
\(1_\mathbb{K}\) is the identity element of multiplication \[\tag{1.5} x\cdot_{\mathbb{K}}1_{\mathbb{K}}=1_{\mathbb{K}}\cdot_{\mathbb{K}}x=x \quad \text{for all }x \in \mathbb{K}.\]
For any \(x \in \mathbb{K}\) there exists a unique element, denoted by \((-x)\) and called the additive inverse of \(x,\) such that \[\tag{1.6} x+_{\mathbb{K}}(-x)=(-x)+_{\mathbb{K}}x=0_\mathbb{K}.\]
For any \(x \in \mathbb{K}\setminus\{0_\mathbb{K}\}\) there exists a unique element, denoted by \(x^{-1}\) or \(\frac{1}{x}\) and called the multiplicative inverse of \(x,\) such that \[\tag{1.7} x\cdot_{\mathbb{K}}\frac{1}{x}=\frac{1}{x}\cdot_{\mathbb{K}}x=1_\mathbb{K}.\]
Distributivity of multiplication over addition \[\tag{1.8} (x+_{\mathbb{K}}y)\cdot_{\mathbb{K}}z=x\cdot_{\mathbb{K}}z+_\mathbb{K}y\cdot_{\mathbb{K}}z \quad \text{for all }x,y,z \in \mathbb{K}.\]
Remark 1.2
It is customary to simply speak of a field \(\mathbb{K},\) without explicitly mentioning \(0_\mathbb{K},1_\mathbb{K}\) and \(+_\mathbb{K},\cdot_{\mathbb{K}}.\)
When \(\mathbb{K}\) is clear from the context, we often simply write \(0\) and \(1\) instead of \(0_{\mathbb{K}}\) and \(1_{\mathbb{K}}.\) Likewise, it is customary to write \(+\) instead of \(+_{\mathbb{K}}\) and \(\cdot\) instead of \(\cdot_{\mathbb{K}}.\) Often \(\cdot_{\mathbb{K}}\) is omitted entirely so that we write \(xy\) instead of \(x\cdot_{\mathbb{K}}y.\)
We refer to the elements of a field as scalars.
The set \(\mathbb{K}\setminus\{0_\mathbb{K}\}\) is usually denoted by \(\mathbb{K}^{*}.\)
For all \(x,y \in \mathbb{K}\) we write \(x-y= x+_{\mathbb{K}}(-y)\) and for all \(x \in \mathbb{K}\) and \(y \in \mathbb{K}^*\) we write \(\frac{x}{y}=x\cdot_{\mathbb{K}}\frac{1}{y}=x\cdot_{\mathbb{K}}y^{-1}.\)
A field \(\mathbb{K}\) containing only finitely many elements is called finite. Algorithms in cryptography are typically based on finite fields.
Example 1.3
The rational numbers or quotients \(\mathbb{Q},\) the real numbers \(\mathbb{R}\) and the complex numbers \(\mathbb{C}\) – that we will study more carefully below – equipped with the usual addition and multiplication are examples of fields.
The integers \(\mathbb{Z}\) (with usual addition and multiplication) are not a field, as only \(1\) and \(-1\) admit a multiplicative inverse.
Considering a set \(\mathbb{F}_2\) consisting of only two elements that we may denote by \(0\) and \(1,\) we define \(+_{\mathbb{F}_2}\) and \(\cdot_{\mathbb{F}_2}\) via the following tables \[\begin{array}{c|c|c}+_{\mathbb{F}_2} & 0 & 1 \\ \hline 0 & 0 & 1 \\ \hline 1 & 1 & 0\end{array} \qquad \text{and} \qquad \begin{array}{c|c|c}\cdot_{\mathbb{F}_2} & 0 & 1 \\ \hline 0 & 0 & 0 \\ \hline 1 & 0 & 1\end{array}\] For instance, we have \(1+_{\mathbb{F}_2}1=0\) and \(1\cdot_{\mathbb{F}_2}1=1.\) Then, one can check that \(\mathbb{F}_2\) equipped with these operations is indeed a field. A way to remember these tables is to think of \(0\) as representing the even numbers, while \(1\) represents the odd numbers. So for instance, a sum of two odd numbers is even and a product of two odd numbers is odd. Alternatively, we may think of \(0\) and \(1\) representing the boolean values FALSE and TRUE. In doing so, \(+_{\mathbb{F}_2}\) corresponds to the logical XOR and \(\cdot_{\mathbb{F}_2}\) corresponds to the logical AND.
Considering a set \(\mathbb{F}_4\) consisting of four elements, say \(\{0,1,a,b\},\) we define \(+_{\mathbb{F}_4}\) and \(\cdot_{\mathbb{F}_4}\) via the following tables \[\begin{array}{c|c|c|c|c}+_{\mathbb{F}_4} & 0 & 1 & a & b \\ \hline 0 & 0 & 1 & a & b \\ \hline 1 & 1 & 0 & b & a \\ \hline a & a & b & 0 & 1 \\ \hline b & b & a & 1 & 0 \end{array}\qquad \text{and} \qquad \begin{array}{c|c|c|c|c}\cdot_{\mathbb{F}_4} & 0 & 1 & a & b \\ \hline 0 & 0 & 0 & 0 & 0 \\ \hline 1 & 0 & 1 & a & b \\ \hline a & 0 & a & b & 1 \\ \hline b & 0 & b & 1 & a \end{array}\] Again one can check that \(\mathbb{F}_4\) equipped with these operations is indeed a field.
Lemma 1.4 • Field properties
In a field \(\mathbb{K}\) we have the following properties:
\(0_{\mathbb{K}}\cdot_{\mathbb{K}} x=0_{\mathbb{K}}\) for all \(x \in \mathbb{K}.\)
\(-x=(-1_\mathbb{K})\cdot_{\mathbb{K}}x\) for all \(x \in \mathbb{K}.\)
For all \(x,y \in \mathbb{K},\) if \(x\cdot_{\mathbb{K}}y=0_\mathbb{K},\) then \(x=0_\mathbb{K}\) or \(y=0_{\mathbb{K}}.\)
\(-0_{\mathbb{K}}=0_{\mathbb{K}}.\)
\((1_{\mathbb{K}})^{-1}=1_{\mathbb{K}}.\)
\((-(-x))=x\) for all \(x \in \mathbb{K}.\)
\((-x)\cdot_{\mathbb{K}}y=x\cdot_{\mathbb{K}}(-y)=-(x\cdot_{\mathbb{K}}y).\)
\((x^{-1})^{-1}=x\) for all \(x \in \mathbb{K}^*.\)
Proof. We will only prove some of the items, the rest are an exercise for the reader.
(i) Using (1.4), we obtain \(0_\mathbb{K}+_{\mathbb{K}}0_{\mathbb{K}}=0_{\mathbb{K}}.\) Hence for all \(x \in \mathbb{K}\) we have \[x\cdot_{\mathbb{K}}0_\mathbb{K}=x\cdot_{\mathbb{K}}(0_\mathbb{K}+0_\mathbb{K})=x\cdot_{\mathbb{K}}0_\mathbb{K}+_{\mathbb{K}}x\cdot_{\mathbb{K}}0_\mathbb{K},\] where the second equality uses (1.8). Adding the additive inverse of \(x\cdot_{\mathbb{K}}0_{\mathbb{K}},\) we get \[x\cdot_{\mathbb{K}}0_{\mathbb{K}}-x\cdot_{\mathbb{K}}0_{\mathbb{K}}=(x\cdot_{\mathbb{K}}0_{\mathbb{K}}+_{\mathbb{K}}x\cdot_{\mathbb{K}}0_{\mathbb{K}})-x\cdot_{\mathbb{K}}0_{\mathbb{K}}\] using the associativity of addition (1.2) and (1.6), this last equation is equivalent to \[0_{\mathbb{K}}=x\cdot_{\mathbb{K}}0_{\mathbb{K}}\] as claimed.
(iii) Let \(x,y \in \mathbb{K}\) such that \(x\cdot_{\mathbb{K}}y=0_{\mathbb{K}}.\) If \(x=0_{\mathbb{K}}\) then we are done, so suppose \(x\neq 0_{\mathbb{K}}.\) Using (1.7), we have \(1_{\mathbb{K}}=x^{-1}\cdot_{\mathbb{K}} x.\) Multiplying this equation with \(y\) we obtain \[y=y\cdot_{\mathbb{K}}1_{\mathbb{K}}=y\cdot_{\mathbb{K}}(x\cdot_{\mathbb{K}} x^{-1})=(y\cdot_\mathbb{K}x)\cdot_{\mathbb{K}} x^{-1}=0_{\mathbb{K}}\cdot_{\mathbb{K}} x^{-1}=0_{\mathbb{K}}\] where we have used (1.5), the commutativity (1.1) and associativity (1.3) of multiplication as well as (i) from above.
(v) By (1.5), we have \(1_{\mathbb{K}}\cdot_{\mathbb{K}}1_{\mathbb{K}}=1_{\mathbb{K}},\) hence \(1_{\mathbb{K}}\) is the multiplicative inverse of \(1_{\mathbb{K}}\) and since the multiplicative inverse is unique, it follows that \((1_{\mathbb{K}})^{-1}=1_{\mathbb{K}}.\)
For a positive integer \(n \in \mathbb{N}\) and an element \(x\) of a field \(\mathbb{K},\) we write \[nx=\underbrace{x+_{\mathbb{K}}x+_{\mathbb{K}}x+_{\mathbb{K}}\cdots+_{\mathbb{K}}x}_{n\text{ summands}}.\] The field \(\mathbb{F}_2\) has the property that \(2x=0\) for all \(x \in \mathbb{F}_{2}.\) In this case we say that \(\mathbb{F}_2\) has characteristic \(2.\) More generally, the smallest positive integer \(p\) such that \(px=0_{\mathbb{K}}\) for all \(x \in \mathbb{K}\) is called the characteristic of the field. In the case where no such integer exists the field is said to have characteristic \(0.\) So \(\mathbb{Q},\mathbb{R},\mathbb{C}\) are fields of characteristic \(0.\) It can be shown that the characteristic of any field is either \(0\) or a prime number.
A subset \(\mathbb{F}\) of a field \(\mathbb{K}\) that is itself a field, when equipped with the multiplication and addition of \(\mathbb{K},\) is called a subfield of \(\mathbb{K}\).
Example 1.5
The rational numbers \(\mathbb{Q}\) form a subfield of the real numbers \(\mathbb{R}.\) Furthermore, as we will see below, the real numbers \(\mathbb{R}\) can be interpreted as a subfield of the complex numbers \(\mathbb{C}.\)
\(\mathbb{F}_2\) may be thought of as the subfield of \(\mathbb{F}_4\) consisting of \(\{0,1\}.\)
1.2 Complex numbers
Historically the complex numbers arose from an interest to make sense of the square root of a negative number. We may picture the rational numbers \(\mathbb{Q}\) as elements of an infinite number line with an origin \(0.\) Positive numbers extending to the right of the origin and negative numbers to the left. Mathematicians have observed early on that this line of numbers contains elements, such as \(\pi\) or \(\sqrt{2},\) that are not quotients. Phrased differently, the rational numbers do not fill out the whole number line, there are gaps consisting of irrational numbers. In a sense to be made precise in the Analysis module, the real numbers may be thought of as the union of the rational numbers and the gaps on the number line, resulting in a gap less line of numbers, known as the complete field of real numbers.
The square \(x^2\) of a real number \(x\) is a non-negative real number, \(x^2\geqslant 0,\) hence if we want to define what the square root of a negative number ought to be, we are in trouble, since there are no numbers left on the line of numbers that we might use. The solution is to consider pairs of real numbers instead. A complex number is an ordered pair \((x,y)\) of real numbers \(x,y \in \mathbb{R}.\) We denote the set of complex numbers by \(\mathbb{C}.\) We equip \(\mathbb{C}\) with the addition defined by the rule \[(x_1,y_1)+_{\mathbb{C}}(x_2,y_2)=(x_1+x_2,y_1+y_2)\] for all \((x_1,y_1)\) and \((x_2,y_2) \in \mathbb{C}\) and where \(+\) on the right denotes the usual addition \(+_{\mathbb{R}}\) of real numbers. Furthermore, we equip \(\mathbb{C}\) with the multiplication defined by the rule \[\tag{1.9} (x_1,y_1)\cdot_{\mathbb{C}}(x_2,y_2)=(x_1\cdot x_2-y_1\cdot y_2,x_1\cdot y_2+y_1\cdot x_2).\] for all \((x_1,y_1)\) and \((x_2,y_2) \in \mathbb{C}\) and where \(\cdot\) on the right denotes the usual multiplication \(\cdot_{\mathbb{R}}\) of real numbers.
Definition 1.6 • Complex numbers
The set \(\mathbb{C}\) together with the operations \(+_{\mathbb{C}},\cdot_{\mathbb{C}}\) and \(0_{\mathbb{C}}=(0,0)\) and \(1_{\mathbb{C}}=(1,0)\) is called the field of complex numbers.
It is sometimes advantageous to think of the real numbers as a subfield of the complex numbers. To make this precise we need the notion of a field embedding:
Definition 1.7 • Field embedding
Let \(\mathbb{F}\) and \(\mathbb{K}\) be fields. A field embedding is a mapping \(\chi : \mathbb{F}\to \mathbb{K}\) satisfying \(\chi(1_\mathbb{F})=1_\mathbb{K},\) \(\chi(0_\mathbb{F})=0_{\mathbb{K}}\) as well as \[\chi(x+_{\mathbb{F}}y)=\chi(x)+_{\mathbb{K}}\chi(y)\qquad \text{and}\qquad \chi(x\cdot_{\mathbb{F}}y)=\chi(x)\cdot_{\mathbb{K}}\chi(y)\] for all \(x,y \in \mathbb{F}.\)
Remark 1.8
It turns out that we don’t have to ask that \(\chi(0_\mathbb{F})=0_{\mathbb{K}}\) in the definition of a field embedding, it is automatically satisfied with the other properties asked for in Definition 1.7. Indeed, we have \[\chi(0_\mathbb{F})=\chi(0_\mathbb{F}+_\mathbb{F}0_\mathbb{F})=\chi(0_\mathbb{F})+_{\mathbb{K}}\chi(0_\mathbb{F}).\] Adding the additive inverse of \(\chi(0_{\mathbb{F}})\) in \(\mathbb{K},\) we conclude that \(0_{\mathbb{K}}=\chi(0_{\mathbb{F}}).\)
A field embedding is injective. Suppose \(x,y \in \mathbb{F}\) satisfy \(\chi(x)=\chi(y)\) so that \(\chi(x-y)=0_{\mathbb{K}}.\) Assume \(w=x-y\neq 0_{\mathbb{F}},\) then \(\chi(w)\cdot_{\mathbb{K}}\chi(w^{-1})=\chi(1_{\mathbb{F}})=1_{\mathbb{K}}.\) Since by assumption \(\chi(w)=0_{\mathbb{K}},\) we thus obtain \(0_{\mathbb{K}}\cdot_{\mathbb{K}}\chi(w^{-1})=1_{\mathbb{K}},\) contradicting Lemma 1.4 (i). It follows that \(x=y\) and hence \(\chi\) is injective.
Example 1.9
From the above tables we see that \(\chi : \mathbb{F}_2 \to \mathbb{F}_4\) defined by \(\chi(1_{\mathbb{F}_2})=1_{\mathbb{F}_4}\) and \(\chi(0_{\mathbb{F}_2})=0_{\mathbb{F}_4}\) is a field embedding.
The mapping \(\chi : \mathbb{R}\to \mathbb{C},\) \(x \mapsto (x,0)\) is a field embedding. Indeed, \[\begin{aligned} \chi(x_1+_\mathbb{R}x_2)&=(x_1+_\mathbb{R}x_2,0)=(x_1,0)+_{\mathbb{C}}(x_2,0)=\chi(x_1)+_{\mathbb{C}}\chi(x_2),\\ \chi(x_1\cdot_\mathbb{R}x_2)&=(x_1\cdot_\mathbb{R}x_2,0)=(x_1,0)\cdot_{\mathbb{C}}(x_2,0)=\chi(x_1)\cdot_{\mathbb{C}}\chi(x_2), \end{aligned}\] for all \(x_1,x_2 \in \mathbb{R}\) and \(\chi(1)=(1,0)=1_{\mathbb{C}}.\)
Item (ii) of the previous example allows to think of the real numbers \(\mathbb{R}\) as the subfield \(\{(x,0)|x \in \mathbb{R}\}\) of the complex numbers \(\mathbb{C}.\) Because of the injectivity of \(\chi,\) it is customary to identify \(x\) with \(\chi(x),\) hence abusing notation, we write \((x,0)=x.\)
Notice that \((0,1)\) satisfies \((0,1)\cdot_{\mathbb{C}}(0,1)=(-1,0)\) and hence is a square root of the real number \((-1,0)=-1.\) The number \((0,1)\) is called the imaginary unit and usually denoted by \(\mathrm{i}.\) Sometimes the notation \(\sqrt{-1}\) is also used. Every complex number \((x,y)\in \mathbb{C}\) can now be written as \[(x,y)=(x,0)+_{\mathbb{C}}(0,y)=(x,0)+_{\mathbb{C}}\mathrm{i}\cdot_{\mathbb{C}}(y,0)=x+\mathrm{i}y,\] where we follow the usual custom of omitting \(\cdot_{\mathbb{C}}\) and writing \(+\) instead of \(+_{\mathbb{C}}\) on the right hand side. With this convention, complex numbers can be manipulated as real numbers, we just need to keep in mind that \(\mathrm{i}\) satisfies \(\mathrm{i}^2=-1.\) For instance, the multiplication of complex numbers \(x_1+\mathrm{i}y_1\) and \(x_2+\mathrm{i}y_2\) gives \[(x_1+\mathrm{i}y_1)(x_2+\mathrm{i}y_2)=x_1x_2+\mathrm{i}^2 y_1y_2+\mathrm{i}(x_1y_2+y_1x_2)=x_1x_2-y_1y_2+\mathrm{i}(x_1y_2+y_1x_2)\] in agreement with (1.9). Here we also follow the usual custom of omitting \(\cdot_{\mathbb{R}}\) on the right hand side.
Definition 1.10
For a complex number \(z=x+\mathrm{i}y \in \mathbb{C}\) with \(x,y \in \mathbb{R}\) we call
\(\operatorname{Re}(z)=x\) its real part;
\(\operatorname{Im}(z)=y\) its imaginary part;
\(\bar z=x-\mathrm{i}y\) the complex conjugate of \(z\);
\(|z|=\sqrt{z\overline{z}}=\sqrt{x^2+y^2}\) the absolute value or modulus of \(z\).
The mapping \(z \mapsto \bar z\) is called complex conjugation.
Remark 1.11
For \(z \in \mathbb{C}\) the following statements are equivalent \[z \in \mathbb{R}\quad \iff \quad \operatorname{Re}(z)=z \quad \iff \quad \operatorname{Im}(z)=0 \quad \iff \quad z=\overline{z}.\]
We have \(|z|=0\) if and only if \(z=0.\)
Example 1.12
Let \(z=\frac{2+5\mathrm{i}}{6-\mathrm{i}}.\) Then \[z=\frac{(2+5\mathrm{i})\overline{(6-\mathrm{i})}}{(6-\mathrm{i})\overline{(6-\mathrm{i})}}=\frac{(2+5\mathrm{i})(6+\mathrm{i})}{|6-\mathrm{i}|^2}=\frac{1}{37}(7+32\mathrm{i}),\] so that \(\operatorname{Re}(z)=\frac{7}{37}\) and \(\operatorname{Im}(z)=\frac{32}{37}.\) Moreover, \[|z|=\sqrt{\left(\frac{7}{37}\right)^2+\left(\frac{32}{37}\right)^2}=\sqrt{\frac{29}{37}}.\]
Remark 1.13
We may think of a complex number \(z=a+\mathrm{i}b\) as a point or a vector in the plane \(\mathbb{R}^2\) with \(x\)-coordinate \(a\) and \(y\)-coordinate \(b.\)
The real numbers form the horizontal coordinate axis (the real axis) and the purely imaginary complex numbers \(\{\mathrm{i}y | y \in \mathbb{R}\}\) form the vertical coordinate axis (the imaginary axis).
The point \(\overline{z}\) is obtained by reflecting \(z\) along the real axis.
\(|z|\) is the distance of \(z\) to the origin \(0_{\mathbb{C}}=(0,0) \in \mathbb{C}\)
The addition of complex numbers corresponds to the usual vector addition.
For the geometric significance of the multiplication, we refer the reader to the Analysis module.
We have the following elementary facts about complex numbers:
Proposition 1.14
For all \(z, w\in \mathbb{C}\) we have
\(\operatorname{Re}(z)=\frac{z+\overline{z}}{2},\) \(\operatorname{Im}(z)=\frac{z-\overline{z}}{2\mathrm{i}}\);
\(\operatorname{Re}(z+w)=\operatorname{Re}(z)+\operatorname{Re}(w),\) \(\operatorname{Im}(z+w)=\operatorname{Im}(z)+\operatorname{Im}(w)\);
\(\overline{z+w}=\overline{z}+\overline{w},\) \(\overline{zw}=\overline{z}\,\overline{w},\) \(\overline{\overline{z}}=z\);
\(|z|^2=|\overline{z}|^2=z\overline{z}=\operatorname{Re}(z)^2+\operatorname{Im}(z)^2\);
\(|zw|=|z||w|.\)
Proof. Exercise.
Exercises
Exercise 1.15
Check that \(\mathbb{C}\) is indeed a field.
Solution
We need to check that all field axioms are verified: Here \(+\) and \(\cdot\) denote addition and multiplication of real numbers respectively and we will use the usual notations \(xy=x\cdot y,\) \(x^2 = x\cdot x,\) \(\frac1x=x^{-1}\) for the multiplicative inverse of \(x\) and also \(x-y\) to denote \(x+(-y).\) Let \((x,y)\in\mathbb{C}\) and \(z_i=(x_i,y_i)\in\mathbb{C}\) for \(i=1,2,3\):
Commutativity of \(+_\mathbb{C}\): Using the commutativity of \(+\) we have \[\begin{aligned} (x_1,y_1)+_\mathbb{C}(x_2,y_2) & = (x_1+x_2,y_1+y_2) = (x_2+x_1,y_2+y_1) \\ & = (x_2,y_2)+_\mathbb{C}(x_1,y_1).\end{aligned}\]
Commutativity of \(\cdot_\mathbb{C}\): Using the commutativity of \(\cdot\) and \(+\) we have \[\begin{aligned}(x_1,y_1)\cdot_\mathbb{C}(x_2,y_2) & = (x_1 x_2-y_1 y_2,x_1 y_2+y_1 x_2) \\ &=(x_2 x_1-y_2 y_1,x_2 y_1+y_2 x_1)\\ & =(x_2,y_2)\cdot_\mathbb{C}(x_1,y_1).\end{aligned}\]
Associativity of \(+_\mathbb{C}\): Using the associativity of \(+\) we have \[\begin{aligned}((x_1,y_1)+_\mathbb{C}(x_2,y_2))+_\mathbb{C}(x_3,y_3) & = (x_1+x_2,y_1+y_2)+_\mathbb{C}(x_3,y_3) \\ & = ((x_1+x_2)+x_3,(y_1+y_2)+y_3) \\ & = (x_1+(x_2+x_3),y_1+(y_2+y_3))\\ & = (x_1,y_1)+_\mathbb{C}((x_2,y_2)+_\mathbb{C}(x_3,y_3)).\end{aligned}\]
Associativity of \(\cdot_\mathbb{C}\): Denoting \(z=(z_1\cdot_\mathbb{C}z_2)\cdot_\mathbb{C}z_3\) we have \[\begin{aligned} z & = (x_1 x_2-y_1 y_2,x_1 y_2+y_1 x_2)\cdot_\mathbb{C}z_3 \\ & = ((x_1 x_2-y_1 y_2)x_3-(x_1 y_2+y_1 x_2)y_3,(x_1 x_2-y_1 y_2)y_3+(x_1 y_2+y_1 x_2)x_3)\\ & = (x_1(x_2x_3-y_2y_3)-y_1(x_2y_3+y_2x_3),x_1(x_2y_3+y_2x_3)+y_1(x_2x_3-y_2y_3))\\ & = z_1\cdot_\mathbb{C}(z_2\cdot_\mathbb{C}z_3). \end{aligned}\]
Identity element of addition: \((x,y) +_\mathbb{C}(0,0) = (x+0,y+0) = (x,y)\)
Identity element of multiplication: \[(x,y) \cdot_\mathbb{C}(1,0) = (x\cdot 1 - y\cdot 0, x\cdot 0 + y\cdot 1) = (x,y)\]
The additive inverse of \((x,y)\) is given by \((-x,-y),\) since \[(x,y)+_\mathbb{C}(-x,-y) = (x+(-x),y+(-y)) = (0,0).\] Suppose \(z\in \mathbb{C}\) had two additive inverses \(z_1,z_2\in\mathbb{C}.\) Then one has \[z_1 = z_1 + 0_\mathbb{C}= z_1 +_\mathbb{C}(z +_\mathbb{C}z_2) = (z_1+_\mathbb{C}z) +_\mathbb{C}z_2 = 0_\mathbb{C}+_\mathbb{C}z_2 = z_2,\] which shows that the additive inverse is unique.
Let \((x,y)\ne(0,0).\) We claim that the multiplicative inverse of \((x,y)\) is given by \(\left(\frac{x}{x^2+y^2},\frac{-y}{x^2+y^2}\right).\) Indeed, we have \[\begin{gathered} (x,y)\cdot_\mathbb{C}\left(\frac{x}{x^2+y^2},\frac{-y}{x^2+y^2}\right)\\ = \left(\frac{x^2}{x^2+y^2}-\frac{y(-y)}{x^2+y^2},\frac{x(-y)}{x^2+y^2}+\frac{x y}{x^2+y^2}\right) = (1,0). \end{gathered}\] Again, the multiplicative inverse is unique: Suppose \(z\in\mathbb{C}\setminus\{0_\mathbb{C}\}\) had the multiplicative inverses \(z_1\) and \(z_2.\) Then \[z_1 = z_1\cdot_\mathbb{C}1_\mathbb{C}= z_1\cdot_\mathbb{C}(z\cdot_\mathbb{C}z_2) = (z_1\cdot_\mathbb{C}z)\cdot_\mathbb{C}z_2 = 1_\mathbb{C}\cdot_\mathbb{C}z_2 = z_2.\]
Distributivity of \(\cdot_\mathbb{C}\) over \(+_\mathbb{C}\): \[\begin{aligned} \left(z_1 +_\mathbb{C}z_2\right)\cdot_\mathbb{C}z_3 & = (x_1+x_2,y_1+y_2)\cdot_\mathbb{C}(x_3,y_3) \\ & = ((x_1+x_2) x_3-(y_1+y_2) y_3,(x_1+x_2) y_3 + x_3 (y_1+y_2))\\ & = (x_1 x_3 - y_1 y_3 + x_2 x_3 -y_2 y_3, x_1 y_3 + y_1 x_3 + x_2 y_3 + y_2 x_3 ) \\ &= (x_1,y_1)\cdot_\mathbb{C}(x_3,y_3) +_\mathbb{C}(x_2,y_2)\cdot_\mathbb{C}(x_3,y_3)\\ & = z_1\cdot_\mathbb{C}z_3 +_\mathbb{C}z_2 \cdot_\mathbb{C}z_3 \end{aligned}\]