Prove that $\mathbb{Q}(i)$ is a subfield of $\mathbb{C}.$

Question

Prove that $\mathbb{Q}(i) := \{a+bi: a,b\in\mathbb{Q}\}$ is a subring of $\mathbb{C}$ (can you check if my proof is correct)?

Prove that $\mathbb{Q}(i)$ is a field (and that it is a subfield of $\mathbb{C}.$)

Side note: just to verify, $\mathbb{Q}(i)$ is NOT an ideal of $\mathbb{C},$ right? However, it is a subring, right?

1) We know $0\in \mathbb{Q}(i)$ so it is nonempty. Let $x\in\mathbb{Q}(i).$ Then $x=a+bi, a,b\in\mathbb{Q}.$ Since $a,b\in\mathbb{Q},$$a,b\in\mathbb{R},$ so $x\in\mathbb{C}.$

Additionally, $\sqrt{2}\in\mathbb{C}$ but $\sqrt{2}\not\in\mathbb{Q}(i),$ so $\emptyset\neq\mathbb{Q}(i)\subsetneq \mathbb{C}.$

Let $x,y\in \mathbb{Q}(i).$ Then $x=a+bi$ and $y=a_2+b_2i,$ where $a,a_2,b,b_2\in\mathbb{Q}.$ Thus $x-y=(a-a_2)+(b-b_2)i\in\mathbb{Q}$ as $(a-a_2)+(b-b_2)\in\mathbb{Q}.$ Additionally, $xy = (aa_2-bb_2)+(ba_2+ab_2)i\in\mathbb{Q}$ as $(aa_2-bb_2),(ba_2+ab_2)\in\mathbb{Q}.$ Therefore $\mathbb{Q}(i)$ is a subring of $\mathbb{C}.$

2) We already proved that $\mathbb{Q}(i)$ is a nonempty subset of $\mathbb{C},$ so we just need to show that it's a subfield. If my understanding is correct, then $\mathbb{Q}(i)$ is unital as $1\in\mathbb{Q}(i)$ and $\forall x\in \mathbb{Q}(i), 1\cdot x=x\cdot 1=x.$ Additionally, we show that $\mathbb{Q}(i)$ is commutative. Let $x,y\in \mathbb{Q}(i).$ Then $x=a+bi$ and $y=a_2+b_2i,$ where $a,a_2,b,b_2\in\mathbb{Q}.$ Thus $xy = (aa_2-bb_2)+(ba_2+ab_2)i$ and $yx=(a_2a-b_2b)+(a_2b+b_2a)i = xy$ by commutativity of multiplication.

Let $x=a+bi\in\mathbb{Q}(i)$ be a nonzero element. Then it has an inverse $\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}i\in\mathbb{Q}(i)$ as $\dfrac{a}{a^2+b^2},\dfrac{b}{a^2+b^2}\in\mathbb{Q}(i).$ Is this enough to show it's a field? I showed it for a general element, so maybe?

Side note: a field (such as $\mathbb C$) has only two ideals — J. W. Tanner, Oct 30 '19 at 21:04
Yes, the two ideals of a field are the trivial ideal and the entire field, because all non-zero field elements are units (invertible), so if an ideal contains a non-zero element then it contains every element of the field — J. W. Tanner, Oct 30 '19 at 21:56
In other words, if an ideal contains a unit, not a "unity", then it is a field? — , Oct 30 '19 at 22:05
$\Bbb Q(i)$ is the fraction field of the ring $\Bbb Z[i]$, hence a field. — Dietrich Burde, Oct 30 '19 at 22:15
@DietrichBurde, in that case I would have to prove that $\mathbb{Z}[i]$ is a field. — , Oct 30 '19 at 22:23
$\mathbb Z[i]$ is a ring, not a field; $\Bbb Q(i)$ is a field — J. W. Tanner, Oct 30 '19 at 22:56
Now we’re getting far afield — pardon the pun — but, yes, $\mathbb Z_3$ is a field and $\mathbb Z_3[i]$ is an extension field of it — J. W. Tanner, Oct 30 '19 at 23:16

score 0 · Accepted Answer · answered Oct 30 '19 at 21:04

No, it is not an ideal, since in a field all ideals are either trivial or the whole field. However, it is a subring (because it satisfies the axioms).

Since you correctly checked all the field axioms, it is indeed enough to do what you did. You even did too much and showed that it is a proper subfield since $\sqrt 2 \notin \mathbb Q(i)$, which I think you were not required to do (but nevertheless is an interesting observation). Congratulations!

Prove that $\mathbb{Q}(i)$ is a subfield of $\mathbb{C}.$

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