proving that $a + b \sqrt {2} + c \sqrt{3} + d \sqrt{6} $ is a subfield of $\mathbb{R}$ The question is given below:



My questions are:
1- How can I find the general form of the multiplicative inverse of each element?
2-How can I find the multiplicative identity?
3-Is the only difference between the field and the subfield definition is that in the case of a subfield every nonzero element has an additive identity but in the field every element not only nonzero ones?
Could anyone help me in understanding these questions, please?
EDIT:
I have found this solution on the internet:



My question: is that a fully acceptable answer to the question? I guess yes.

Let $F$ be a sub-field of the real numbers $\Bbb R$. If for some integer $k \gt 1$
$\quad \sqrt k \notin F$
then using elementary algebra it can be shown that the set
$\tag 1 \{ s_1 + s_2 \sqrt {k} \;\; | \, s_1,s_2 \in F\}$
is a field and that
$$\tag 2 ( s_1 + s_2 \sqrt {k})^{-1} = \frac{s_1 - s_2 \sqrt k}{s_1^2 - ks_2^2}$$
Since
$\tag 3 a + b \sqrt {2} + c \sqrt{3} + d \sqrt{6} = (a + b \sqrt {2}) + (c + d \sqrt{2}) \sqrt 3$
the OP is indeed dealing with a subfield of $\Bbb R$.
To arrive at a general formula for the inverse of $a + b \sqrt {2} + c \sqrt{3} + d\sqrt{6}$ in the form $a' + b' \sqrt {2} + c' \sqrt{3} + d'\sqrt{6}$ is unwieldy but can be done by applying (when necessary) $\text{(2)}$ twice.
Since the OP's question has the linear algebra tag, note that the general form for the multiplicative inverse is given by applying the inverse of
$$
\begin{bmatrix}a & 2b & 3c & 6d\\ b & a & 3d & 3c \\ c & 2d & a & 2b\\d & c & b & a\end{bmatrix}
$$
to the vector
$$
\begin{bmatrix}1 \\ 0 \\ 0 \\0 \end{bmatrix}
$$
to calculate
$$
\begin{bmatrix}a' \\ b' \\ c' \\d' \end{bmatrix}
$$
ANSWER:
$\frac{1}{a^4 - 2 a^2 (2 b^2 + 3 c^2 + 6 d^2) + 48 a b c d + 4 b^4 - 12 b^2 (c^2 + 2 d^2) + 9 (c^2 - 2 d^2)^2} \times$
\begin{bmatrix}a^3 - 2 b^2 a - 3 c^2 a - 6 d^2 a + 12 b c d \\2 b^3 - a^2 b - 3 c^2 b - 6 d^2 b + 6 a c d \\ 3 c^3 - a^2 c - 2 b^2 c - 6 d^2 c + 4 a b d   \\6 d^3 - a^2 d - 2 b^2 d - 3 c^2 d + 2 a b c   \end{bmatrix}
