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.

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.)
