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.
 A: Regarding 1: Note that if we multiple your general element by its conjugate with respect to $\sqrt{3}$,
$$  (a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6})(a + b\sqrt{2} - c\sqrt{3} - d\sqrt{6}) \\ = a^2 + 2b^2 - 3c^2 -6d^2 + (2ab-6cd)\sqrt{2}  \text{,}  $$
which we can simplify further by multiplying with this new expression's conjugate with respect to $\sqrt{2}$, $a^2 + 2b^2 - 3c^2 -6d^2 - (2ab-6cd)\sqrt{2}$.  We find
$$   (a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6})(a + b\sqrt{2} - c\sqrt{3} - d\sqrt{6})(a^2 + 2b^2 - 3c^2 -6d^2 - (2ab-6cd)\sqrt{2})  \\
= a^4-4 a^2 b^2-6 a^2 c^2-12 a^2 d^2+48 a b c d+4 b^4-12 b^2 c^2-24 b^2 d^2+9 c^4-36c^2 d^2+36 d^4  \text{.}  $$  Now divide both sides by $(a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6})$ and $(a^4-4 a^2 b^2-6 a^2 c^2-12 a^2 d^2+48 a b c d+4 b^4-12 b^2 c^2-24 b^2 d^2+9 c^4-36c^2 d^2+36 d^4)$ to find an expression for the multiplicative inverse of $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$.
Notice that this is entirely analogous to what we do with complex numbers.  For $a+b\mathrm{i}$, we multiply by its conjugate (with respect to $\mathrm{i}$), 
$$  (a + b\mathrm{i})(a - b\mathrm{i}) = a^2 + b^2  \text{.}  $$
Then $\frac{a - b \mathrm{i}}{a^2 + b^2} = \frac{1}{a + b \mathrm{i}}$.
Regarding 2: What's the multiplicative identity in $\mathbb{R}$?  Is it possible for the multiplicative identity of a subfield to be a different number?  Suppose $s$ in a subfield satisfies $sx = x$ for every $x$ in that subfield, then that equation is also true for all those elements in the field.  If we write $1 \in \mathbb{R}$ as the multiplicative identity in $\mathbb{R}$, then $sx = 1x$ is an equivalent equation in $\mathbb{R}$ and $(s-1)x = 0$.  It is a familiar property of $\mathbb{R}$ that this forces either $s-1 = 0$ or $x = 0$.  Since this holds for at least one non-zero $x$ (because subfields are fields, so they have at least two elements), it must be the case that $s-1 = 0$, so $s = 1$.
Regarding 3:  Every element in any field (or subfield) has an additive inverse.  For a field $F$ and an element $x \in F$, its additive inverse is $-x$.  For your element $a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}$, its additive inverse is $-a -b\sqrt{2} -c\sqrt{3} -d\sqrt{6}$ for every choice of $a,b,c,d$.  Even zero has an additive inverse.  (It's zero.)
A: Here it is for $a+b\sqrt{2}$.
The multiplicative identity is $1$, or you could write it as $1+0\sqrt2$.
A subfield is the same as a field.  You just have to check the sum, difference, product and ratio stays in the set.
To check the ratio stays in the set, I just do the multiplicative inverse.  That and multiplication gives any ratio.
Suppose the inverse of $a+b\sqrt2$ is $c+d\sqrt2$.  Then
$$(a+b\sqrt2)(c+d\sqrt2)=1\\
ac+2bd+(ad+bc)\sqrt2=1+0\sqrt2\\
ac+2bd=1,ad+bc=0$$
The last two equations can be written as a matrix equation
$$\left[\begin{array}{cc}a&2b\\b&a\end{array}\right]
\left[\begin{array}{c}c\\d\end{array}\right]=
\left[\begin{array}{c}1\\0\end{array}\right]$$
The matrix's determinant is $a^2-2b^2$.  If its determinant is zero for nonzero rationals $a$ and $b$, then $\sqrt2=a/b$ is rational.  So the determinant is nonzero and you can invert the matrix to find $c$ and $d$.
You will have a $4×4$ matrix instead of a $2×2$.  
A: I think that all will agree that the only problem is deciding that a number of form $a+b\sqrt2+c\sqrt3+d\sqrt6$ (with $a,b,c,d\in\Bbb Q$) has a reciprocal that’s again of this form. You don’t need to write out the reciprocal, just give an adequate argument that your number has one. Let’s do it this way:
$$
\begin{align}
\frac1{a+b\sqrt2+c\sqrt3+d\sqrt6}&=\frac1{(a+b\sqrt2\,)+(c+d\sqrt2\,)\sqrt3}\\
\\
&=\frac{(a+b\sqrt2\,)-(c+d\sqrt2\,)\sqrt3}{(a+b\sqrt2\,)^2-3(c+d\sqrt2\,)^2}\,,
\end{align}
$$
in which the numerator is of form $a'+b'\sqrt2+c'\sqrt3+d'\sqrt6$, and the denominator now has the form $A+B\sqrt2$, for suitable rational numbers $A$ and $B$. But you already know, from high-school, how to find
$1/(A+B\sqrt2)$.
And there you are.
A: Check the field axioms for such expressions. You really need only that the sum and product has the same form, identify the additive and multiplicative inverses; the others follow as you are operating on real numbers. They hold, and the set is clearly a (proper) subset of $\mathbb{R}$. Done.
A: 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}
