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mathmari

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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)