Dimension of Vector Space

Determine whether each of the following is a vector space and find the dimension and a basis for each that is a vector space:
i) $\mathbb{C}$ over $\mathbb{C}$.
ii) $\mathbb{C}$ over $\mathbb{R}$.
iii) $\mathbb{R}$ over $\mathbb{C}$.
iv) $\mathbb{R}$ over $\mathbb{Q}$.
v) $\mathbb{Q}$ over $\mathbb{R}$.
vi) $\mathbb{Q}$ over $\mathbb{Z}$.
vii) $\mathbb{S}=\{a+b\sqrt{2}+c\sqrt{5}\:|\:a,b,c \in \mathbb{Q}\}$ over $\mathbb{Q,R,\text{or}\: C}$.

Answers:
i) Yes. $\{1\}$ is a basis. Dimension is $1$.
[Say $1+2i,1+i\in \Bbb C$. $(1+2i)\cdot (1+i)=-1+3i=1\cdot (-1+3i)$, where $1$ is the basis element and $-1+3i$ is in the field $\Bbb C$.]

ii) Yes. $\{1,i\}$ is a basis. Dimension is $2$.
[Say $1+i\in \Bbb C$ and $2\in \Bbb R$. $2\cdot (1+i)=2+2i=2\cdot 1+2\cdot i$, where $1,i$ are elements in the basis and $2$ is in the field $\Bbb R$.]

iii) No. $i\in \Bbb C$ and $1\in \Bbb R$, but $i \cdot 1=i\notin \Bbb R$.
iv) Yes. $\{1,\pi,\pi^2,\cdots\}$ are linearly independent over $\Bbb Q$. Dimension is infinite.
v) No. $\sqrt{2}\in \Bbb R$ and $1\in \Bbb Q$, but $\sqrt{2} \cdot 1=\sqrt{2}\notin \Bbb Q$.
vi) No. $\Bbb Z$ is not a field.
vii) Yes only over $\Bbb Q$. $\{1,\sqrt{2},\sqrt{5}\}$ is a basis. Dimension is $3$.

From iii and v, we know that the field is always 'smaller' than its vector space. In fact, this is related to the concept of field extension (to be discussed in our next post).

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Let $V=\{(x,y)\:|\:x,y\in \Bbb C\}$. Under the standard addition and scalar multiplication for ordered pairs of complex numbers, is $V$ a vector space over $\Bbb C$? Over $\Bbb R$? Over $\Bbb Q$? If so, find the dimension of $V$.

Answers:
From the previous question, we know $\{(1,0),(0,1)\}$ form a basis for $V$ over $\Bbb C$. Dimension is $2$. $\{(1,0),(i,0),(0,1),(0,i)\}$ form a basis for $V$ over $\Bbb R$. Dimension is $4$. Lastly, the dimension of $V$ over $\Bbb Q$ is infinite.