4.1 Construction of $\mbox{{\bf C}}_F$.

Throughout this chapter, $F$ will represent a field in which $-1$ is not a square. For example, in an ordered field $-1$ is not a square, but in $\mbox{{\bf Z}}_5$, $(2)^2=4=-1$ so $-1$ is a square. In $\mbox{{\bf Z}}_3$,

$$0^2=0,\;\; 1^2=1,\;\; 2^2=1, \; \mbox{ and } -1=2,$$

so $-1$ is not a square in $\mbox{{\bf Z}}_3$.

Let $F$ be a field in which $-1$ is not a square. I am going to construct a new field $\mbox{{\bf C}}_F$ which contains (a copy of) $F$ and a new element $i$ such that $i^2 = -1$. The elements of $\mbox{{\bf C}}_F$ will all have the form

$$a+bi$$

where $a$ and $b$ are in $F$. I'll call $\mbox{{\bf C}}_F$ the complexification of $F$. Before I start my construction, note that if $a,b,c,d$ are in $F$ and $i^2 = -1$, then by the usual field axioms

$$(a+bi)+(c+di)=(a+c)+(b+d)i,$$ (4.1)

and

$$(a+bi)(c+di)=(ac-bd)+(ad+bc)i.$$ (4.2)

Let $F$ be a field in which $-1$ is not a square. Let $\mbox{{\bf C}}_F=F\times F$ denote the Cartesian product of $F$ with itself (Cf. definition 1.55). I define two binary operations $\oplus$ and $\odot$ $\mbox{{\bf C}}_F$ as follows (cf. (4.1) and (4.2)): for all $(a,b),(c,d)\in\mbox{{\bf C}}_F$,

$$(a,b)\oplus (c,d)=(a+c,b+d)$$

and

$$(a,b)\odot (c,d)=(ac-bd,ad+bc).$$

We will now show that $(\mbox{{\bf C}}_F,\oplus,\odot)$ is a field.

4.3   Theorem (Associativity of $\odot$.) The operation $\odot$ is associative on $\mbox{{\bf C}}_F$.

Proof: Let $(a,b),(c,d)$ and $(e,f)$ be elements in $\mbox{{\bf C}}_F$. Then

$${ (a,b)\odot\left((c,d)\odot(e,f)\right)} \hspace{1cm}$$

$$= (a,b)\cdot(ce-df,cf+de)$$

$$= \left(a(ce-df)-b(cf+de), a(cf+de)+b(ce-df)\right)$$

$$= (ace-adf-bcf-bde,acf+ade+bce-bdf).$$ (4.4)

Also,

$${((a,b) \odot (c,d))\odot(e,f)} \hspace{1cm}$$

$$= (ac-bd,ad+bc)\odot(e,f)$$

$$= \left((ac-bd)e-(ad+bc)f, (ac-bd)f+(ad+bc)e\right)$$

$$= (ace-bde-adf-bcf, acf-bdf+ade+bce).$$ (4.5)

Now by using the field properties of $F$, we see that the (4.4) and (4.5) are equal, and hence

$$(a,b)\odot\left((c,d)\odot(e,f)\right)=\left((a,b)\odot(c,d)\right)\odot(e,f).$$

Hence, $\odot$ is associative on $\mbox{{\bf C}}_F$. $\mid\!\mid\!\mid$

I expect the multiplicative identity for $\mbox{{\bf C}}_F$ to be $1+0i=(1,0)$.

4.6   Theorem (Multiplicative identity for $\mbox{{\bf C}}_F$.)The element $(1,0)$ is an identity for $\odot$ on $\mbox{{\bf C}}_F$.

Proof: For all $(a,b)\in\mbox{{\bf C}}_F$, we have

$$(1,0)\odot(a,b)=(1\cdot a-0\cdot b,1\cdot b+0\cdot a)=(a,b)$$

and

$$(a,b)\odot(1,0)=(a\cdot 1-b\cdot 0,a\cdot 0+b\cdot 1)=(a,b).\mbox{ $\mid\!\mid\!\mid$}$$

4.7   Exercise.

a)

Show that $\oplus$ is associative on $\mbox{{\bf C}}_F$.

b)

Show that there is an identity for $\oplus$ on $\mbox{{\bf C}}_F$.

c)

Show that every element in $\mbox{{\bf C}}_F$ has an inverse for $\oplus$.

d)

Show that $\odot$ is commutative on $\mbox{{\bf C}}_F$.

e)

Show that the distributive law holds for $\mbox{{\bf C}}_F$.

f)

Show that the additive and multiplicative identities for $\mbox{{\bf C}}_F$ are different.

As a result of exercise 4.7 and the two previous theorems, we have verified that $(\mbox{{\bf C}}_F,\oplus,\odot)$ satisfies all of the field axioms except existence of multiplicative inverses. Note that up to this point we have never used the assumption that $-1$ is not a square in $F$.

4.8   Theorem (Existence of multiplicative inverses.)Let $F$ be a field in which $-1$ is not a square and let $(a,b)$ be an element in $\mbox{{\bf C}}_F\backslash\{(0,0)\}$. Then $(a,b)$ has an inverse for $\odot$.

Proof: Let $(a,b)\in\mbox{{\bf C}}_F\backslash\{(0,0\}$. I want to find a point $(x,y)\in\mbox{{\bf C}}_F$ such that

$$(a,b)\odot (x,y)=(1,0).$$

Since multiplication is commutative, this shows that $(x,y)\odot(a,b)=(1,0)$ and hence that $(x,y)$ is a multiplicative inverse for $(a,b)$. I want

$$(ax-by,ay+bx)=(1,0),$$

so I want

$$bx+ay=0$$ (4.9)

and

$$ax-by=1.$$ (4.10)

Multiply the first equation by $b$ and the second by $a$ to get

$$\begin{eqnarray*} b^2 x+aby&=&0 \\ a^2 x-aby&=&a. \end{eqnarray*}$$

If we add these equations, we get

$$(a^2+b^2)x=a.$$ (4.11)

In the next lemma I'll show that if $-1$ is not a square then $a^2+b^2\neq 0$ for all $(a,b)\in\mbox{{\bf C}}_F\backslash\{(0,0)\}$, so by (4.11), $${x={a\over {a^2+b^2}}}$$. Now multiply (4.9) by $a$ and (4.10) by $-b$ to get

$$\begin{eqnarray*} abx+a^2y&=&0 \\ -abx+b^2y&=&-b. \end{eqnarray*}$$

If we add these equations, we get

$$(a^2+b^2)y=-b$$

so

$$y={{-b}\over {a^2+b^2}}.$$

I've shown that if $(a,b)\odot(x,y)=(1,0)$, then $${(x,y)=\left({a\over {a^2+b^2}},{{-b}\over {a^2+b^2}}\right)}$$. A direct calculation shows that this works:

$$(a,b)\odot\left( {a\over {a^2+b^2}},{{-b}\over {a^2+b^2}}\ri... ...ba}\over {a^2+b^2}}\right) =(1,0).\mbox{ $\mid\!\mid\!\mid$}$$

4.12   Remark. The above proof shows that for all $(a,b) \in\mbox{{\bf C}}\setminus\{(0,0)\}$,

$$(a,b)^{-1} = \left( {a\over a^2+b^2},{-b\over a^2+b^2}\right).$$

4.13   Lemma. Let $F$ be a field in which $-1$ is not a square. Let $(a,b)$ be an element in $\mbox{{\bf C}}_F\backslash\{(0,0)\}$. Then $a^2+b^2\neq 0$.

Proof: Since $(a,b)\neq(0,0)$, either $a\neq 0$ or $b\neq 0$.

Case 1:

Suppose $a\neq 0$, then $a^2\neq 0$, so

$$\begin{eqnarray*} a^2+b^2=0&\mbox{$\Longrightarrow$}&a^2\left(1+\left({b\over a}... ...^2=0 \\ &\mbox{$\Longrightarrow$}&\left({b\over a}\right)^2=-1. \end{eqnarray*}$$

Since $-1$ is not a square in $F$, $a^2+b^2\neq 0$.

Case 2:

Suppose $b\neq 0$. Repeat the argument of Case 1 with the roles of $a$ and $b$ interchanged. $\mid\!\mid\!\mid$

We now have verified all of the field axioms so we know that $\mbox{{\bf C}}_F$ is a field. Hence we can calculate in $\mbox{{\bf C}}_F$ using all of the algebraic results that have been proved to hold in all fields.

4.14   Notation ($i,\tilde a$) Let $F$ be a field in which $-1$ is not a square. We will denote the pair $(0,1)\in\mbox{{\bf C}}_F$ by $i$, and if $a\in F$ we will denote the pair $(a,0)\in\mbox{{\bf C}}_F$ by $\tilde a$.

We have $\tilde 0=(0,0)$ is the additive identity for $\mbox{{\bf C}}_F$, and $\tilde 1=(1,0)$ is the multiplicative identity for $\mbox{{\bf C}}_F$. If $a\in F$, then $-\tilde a=-(a,0)=(-a,0)=\widetilde{-a}$. Also

$$i^2=(0,1)\odot(0,1)=(0-1,0)=(-1,0)=-\tilde 1,$$

so $i$ is a square root of $-\tilde 1$.

If $a,b\in F$, then

$$\begin{eqnarray*} \tilde a \oplus (\tilde b \odot i) &=&(a,0) \oplus ((b,0)\odot (0,1)) \\ &=&(a,0) \oplus (0,b)=(a,b), \end{eqnarray*}$$

and hence every element $(a,b)\in\mbox{{\bf C}}_F$ can be written in the form $\tilde a \oplus (\tilde b\odot i)$. We have

$$\begin{eqnarray*} \tilde a\odot\tilde b&=&(a,0)\odot(b,0)=(ab,0)=\widetilde{ab} ... ...ilde a\oplus\tilde b&=&(a,0)\oplus(b,0)=(a+b,0)=\widetilde{a+b}. \end{eqnarray*}$$

Hence $\mbox{{\bf C}}_F$ contains a `` copy of $F$". Each element $a$ in $F$ corresponds to a unique $\tilde a$ in $\mbox{{\bf C}}_F$ in such a way that addition in $\mbox{{\bf C}}_F$ corresponds to addition in $F$ and multiplication in $\mbox{{\bf C}}_F$ corresponds to multiplication in $F$. We will henceforth drop the tildes, and we'll denote $\oplus$ by $+$ and $\odot$ by $\cdot$ as is usual in fields. Then every element in $\mbox{{\bf C}}_F$ can be written uniquely as $a+bi$ where $a,b\in F$ and $i^2 = -1$.

We consider $F$ to be a subset of $\mbox{{\bf C}}_F$. An element $z=(a,b)=a+bi$ of $\mbox{{\bf C}}_F$ is in $F$ if and only if $b=0$. If $a,b,c,d\in F$, then

$$a+bi=c+di\hspace{1ex}\Longleftrightarrow\hspace{1ex}(a,b)=(c,d)\hspace{1ex}\Longleftrightarrow\hspace{1ex}a=c \mbox{ and } b=d.$$

4.15   Examples. I will find the square roots of $2i$ in $\mbox{{\bf C}}_{\mathbf Q}$. Let $a,b\in\mbox{{\bf Q}}$. Then

$$\begin{eqnarray*} (a+bi)^2=2i&\mbox{$\Longleftrightarrow$}&a^2-b^2+2abi=2i \\ &... ...$}&(a=b \mbox{ and } ab=1) \mbox{ or } (a=-b \mbox{ and } ab=1). \end{eqnarray*}$$

Now

$$(a=b \mbox{ and } ab=1)\hspace{1ex}\Longleftrightarrow\hspace... ... a^2=1)\hspace{1ex}\Longleftrightarrow\hspace{1ex}a+bi=\pm(1+i)$$

and

$$(a=-b\mbox{ and } ab=1)\mbox{$\hspace{1ex}\Longrightarrow\hsp... ...}$}-b^2=1\mbox{$\hspace{1ex}\Longrightarrow\hspace{1ex}$}b^2=-1$$

which is impossible. The only possible square roots of $2i$ are $\pm (1+i)$. You can easily verify that these are square roots of $2i$.

4.16   Example. I can solve the quadratic equation

$$z^2-4z+4-{1\over 2}i=0$$ (4.17)

in $\mbox{{\bf C}}_{\mathbf Q}$ by using the quadratic formula for $Az^2+Bz+C=0$. $${B^2-4AC=16-4\left(4-{1\over 2}i\right)=2i=(1+i)^2}$$ (by the previous example). Since $B^2-4AC$ is a square, the equation has the solution set

$$\left\{ {{4+(1+i)}\over 2},{{4-(1+i)}\over 2}\right\}=\left\{ {5\over 2}+{1\over 2}i,{3\over 2}-{1\over 2}i\right\}.$$

4.18   Exercise. Check that $${ {5\over 2}+{1\over 2}i}$$ and $${ {3\over 2}-{1\over 2}i}$$ are solutions to (4.17).

4.19   Exercise. A

a)

Write $${ {1\over {1-2i}}}$$ in the form $a+bi$ where $a,b\in\mbox{{\bf Q}}$.

b)

Find all solutions to

$$(1-2i)z^2-2z+1=0$$

in $\mbox{{\bf C}}_{\mathbf{Q}}$. (You may want to use the result of example 4.15.) Write your solutions in the form $a+bi$ where $a,b\in\mbox{{\bf Q}}$.

4.20   Entertainment. We noted earlier that $-1$ is not a square in $\mbox{{\bf Z}}_3$, so $\mbox{{\bf Z}}_3$ has a complexification, which is a field with 9 elements. Show that if $z=1+i$, then the 9 elements in $\mbox{{\bf C}}_{\mathbf{Z}_3}$ are

$$\{0,z,z^2,z^3,z^4,z^5,z^6,z^7,z^8\}.$$

Can you figure out before you make any calculations which of these elements is $1$?