Smallest subfield of C that contains i

[mathmari]

Hey!

I am looking at the following:
Determine the smallest subfield $\mathbb{K}$ of $\mathbb{C}$ that contains the complex number $i$.

The smallest subfield of $\mathbb{C}$ that contains the complex number $i$ is $\mathbb{K}=\mathbb{Q}(i)$, right?
Do we have to show that $\mathbb{Q}(i)=\{a+bi\mid a,b\in \mathbb{Q}\}$ is a field and then that it is the smallest subfield of $\mathbb{C}$ that contains $i$ ?

For the first part, we have to show that $\mathbb{Q}(i)$ is closed under addition and multiplication, contains both additive and multiplicative identities, contains additive inverses, and contains multiplicative inverses for all nonzero elements, right?

• $\mathbb{Q}(i)$ is closed under addition :
  
  Let $x,y\in \mathbb{Q}(i)$, so $x=a_1+b_1i$ and $y=a_2+b_2i$. Then $x+y=(a_1+a_2)+(b_1+b_2)i\in \mathbb{Q}(i)$.
  
• $\mathbb{Q}(i)$ is closed under multiplication :
  
  Let $x,y\in \mathbb{Q}(i)$, so $x=a_1+b_1i$ and $y=a_2+b_2i$. Then $x\cdot y=a_1b_1+a_1b_2i+a_2b_1i-a_2b_2=(a_1b_1-a_2b_2)+(a_1b_2+a_2b_1)i\in \mathbb{Q}(i)$.
  
• $\mathbb{Q}(i)$ contains additive identity :
  
  The additive identity is $0 = 0+0i$.
  
• $\mathbb{Q}(i)$ contains multiplicative identity :
  
  The multiplicative identity is $1 = 1+0i$.
  
• $\mathbb{Q}(i)$ contains additive inverses
  
  Let $x\in \mathbb{Q}(i)$, so $x=a+bi$. Then the additive inverse is $-x=-a-bi$.
  
• $\mathbb{Q}(i)$ contains multiplicative inverses for all nonzero elements
  
  Let $x\in \mathbb{Q}(i)$, so $x=a+bi$. Then the multiplicative inverse is $\frac{1}{x}=\frac{1}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i\in \mathbb{Q}(i)$.
  

So, $\mathbb{Q}(i)$ is a field. For the second part:
Any field $F$ that contains $\mathbb Q$ and $i$ also contains $qi$ with $q$ rational, since $F$ is closed under products and $q$ and $i$ belong to $F$. Since $F$ contains every rational $p$ and every number of the form $qi$ it contains also their sum $p+qi$.

This proves $Q(i)\subseteq F$, which means that $\mathbb{Q}(i)$ is the smallest subfield of $\mathbb{C}$ that contains $i$. Is everything correct? Could I improve something? (Wondering)

 

[I like Serena]

For the second part:
Any field $F$ that contains $\mathbb Q$ and $i$ also contains $qi$ with $q$ rational, since $F$ is closed under products and $q$ and $i$ belong to $F$. Since $F$ contains every rational $p$ and every number of the form $qi$ it contains also their sum $p+qi$.

This proves $Q(i)\subseteq F$, which means that $\mathbb{Q}(i)$ is the smallest subfield of $\mathbb{C}$ that contains $i$.

Is everything correct? Could I improve something? (Wondering)

Hey mathmari! (Smile)

The first part looks fine to me, although we may want to mention for the multiplicative inverse that $a^2+b^2\ne 0$ so that the inverse is well defined for all $x\in \mathbb Q(i) \setminus \{0\}$.

As for the second part, it seems we're already assuming that $\mathbb Q$ must be contained in the sub field.
It's true alright, but why? (Wondering)

 

[mathmari]

The first part looks fine to me, although we may want to mention for the multiplicative inverse that $a^2+b^2\ne 0$ so that the inverse is well defined for all $x\in \mathbb Q(i) \setminus \{0\}$.

Ah ok!

As for the second part, it seems we're already assuming that $\mathbb Q$ must be contained in the sub field.
It's true alright, but why? (Wondering)

So, we have to show that each subfield of the complex numbers contains every rational number, right?

Let $F$ be a subfield of $\mathbb{C}$.

Since $F$ is a field it must contain a distinguished element $\tilde{0}$, the additive identity. Since $F$ is a subfield of $\mathbb{C}$, we have that $F\subseteq \mathbb{C}$, so $\tilde{0}$ will be also the additive identity in $\mathbb{C}$. But this element is unique, therefore $\tilde{0} = 0$. Therefore $0 \in F$.

Since $F$ is a field it must contain a distinguished element $\tilde{1}$, the multiplicative identity. Since $F$ is a subfield of $\mathbb{C}$, we have that $F\subseteq \mathbb{C}$, so $\tilde{1}$ will be also the multiplicative identity in $\mathbb{C}$. But this element is unique, therefore $\tilde{1} = 1$. Therefore $1 \in F$.

Since $F$ is a field, it is closed under addition. We will show by induction that $\mathbb{N}\subseteq F$.
Base case: Since $1 \in F$ then we must have that $1 + 1 = 2 \in F$.
Inductive hypothesis: We suppose that $n\in F$.
Inductive step: Since $n\in F$ and $1\in F$ and since $F$ is closed under addition we get that $n+1\in F$.
So, we have that $n\in F$ for each $n\in \mathbb{N}$. This means that $\mathbb{N}\subseteq F$.

Since $F$ is a filed, it must contain all the additive inverses of all its elements. We have that $n\in \mathbb{N}\subseteq F$ so it must also $-n\in F$. So since $\mathbb{N} \subseteq F$ we must have $-\mathbb{N}=\mathbb{Z} \subseteq F$.

Since $F$ is a field, it must also contain all multiplicative inverses of its non-zero elements, therefore $\frac{1}{\mathbb{Z}}:=\left \{\frac{1}{n}\mid n\in \mathbb{Z}\right \}\subseteq F$.
Let $\frac{a}{b}$ be a rational number, with $a, b\in \mathbb{Z}$ and $b\neq 0$.
We have that $a\in \mathbb{Z}\subseteq F$ and $\frac{1}{b}\in \frac{1}{\mathbb{Z}}\subseteq F$.
Since $F$ is a field, it is closed under multiplication, we get that $\frac{a}{b}=a\cdot \frac{1}{b}\in F$.
Therefore, $F$ contains every rational number, i.e. $\mathbb{Q}\subseteq F$. Is everything correct? (Wondering)

 

[I like Serena]

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

Thank you! (Bow)