# Smallest subfield of C that contains i

• MHB
• mathmari
In summary, the conversation discusses determining the smallest subfield of the complex numbers that contains the complex number $i$. The smallest subfield is identified as $\mathbb{K}=\mathbb{Q}(i)$ and it is shown that it is a field. The second part of the conversation verifies that any subfield of the complex numbers must contain all rational numbers, and this is proven by showing that $0, 1, \mathbb{N}, \mathbb{Z},$ and $\mathbb{Q}$ are all contained in the subfield $\mathbb{K}$.
mathmari
Gold Member
MHB
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)

mathmari said:
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)

I like Serena said:
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!

I like Serena said:
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)

Yep. (Nod)

I like Serena said:
Yep. (Nod)

Thank you! (Bow)

## 1. What is the smallest subfield of C that contains i?

The smallest subfield of C that contains i is the complex numbers, denoted by C(i). Complex numbers are numbers that can be expressed in the form a + bi, where a and b are real numbers and i is the imaginary unit (√-1).

## 2. How is the smallest subfield of C that contains i defined?

The smallest subfield of C that contains i is defined as the set of all numbers that can be written as a + bi, where a and b are real numbers and i is the imaginary unit. This set includes all real numbers and imaginary numbers, as well as their combinations.

## 3. Can you give an example of a number in the smallest subfield of C that contains i?

One example of a number in the smallest subfield of C that contains i is 2 + 3i. This number is a combination of a real number (2) and an imaginary number (3i). Other examples include -5i, 1 + 2i, and √2 + 4i.

## 4. What is the significance of the smallest subfield of C that contains i?

The smallest subfield of C that contains i is significant because it allows for the representation and manipulation of complex numbers. These numbers are used in various fields of science and engineering to model and solve real-world problems that involve both real and imaginary quantities.

## 5. How is the smallest subfield of C that contains i related to the complex plane?

The smallest subfield of C that contains i is closely related to the complex plane, which is a geometric representation of complex numbers. The real numbers are plotted on the horizontal axis, while the imaginary numbers are plotted on the vertical axis. This allows for the visualization and understanding of complex numbers and their operations, such as addition, subtraction, multiplication, and division.

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