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Why is $K:=\{x+\sqrt{2}y : x,y\in \mathbb{Q}\}$ a field?

I know that if $K$ is a sub-field of a field (in this case it should be $\mathbb{R}$?), it is a field itself.

Firstly, have to prove that if $a,b\in K \Longrightarrow a+b\in K$ $\land$ $a\cdot b \in K$

I am having trouble figuring out what $a$ and $b$ are since $x+\sqrt{2}y$ is an addition... any suggestions?

Analysis
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4 Answers4

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If

$K = \{x + y\sqrt 2, \; x, y \in \Bbb Q \} \tag 1$

and

$a \in K, \tag 2$

then there exist

$a_1, a_2 \in \Bbb Q \tag 3$

such that

$a = a_1 + a_2\sqrt 2; \tag 4$

likewise,

$b \in K \tag 5$

takes the form

$b = b_1 + b_2\sqrt 2, \; b_1, b_2 \in \Bbb Q; \tag 6$

then

$a + b = (a_1 + a_2\sqrt 2) + (b_1 + b_2\sqrt 2) = (a_1 + b_1) + (a_2 + b_2)\sqrt 2 \in K \tag 7$

since

$a_1 + b_1, a_2 + b_2 \in \Bbb Q; \tag 8$

also,

$ab = (a_1 + a_2\sqrt 2)(b_1 + b_2\sqrt 2) = a_1b_1 + a_1b_2\sqrt 2 + a_2b_1\sqrt 2 + a_2b_2(\sqrt 2)^2$ $= a_1b_1 + a_1b_2\sqrt 2 + a_2b_1\sqrt 2 + 2a_2b_2 = (a_1b_1 + 2a_2b_2) + (a_1b_2+ a_2b_1)\sqrt 2 \in K, \tag 9$

since again

$a_1b_1 + 2a_2b_2, a_1b_2+ a_2b_1 \in \Bbb Q, \tag{10}$

valid since $\Bbb Q$ is itself a field, hence closed under addition and multiplication. Clearly

$0 = 0 + 0(-\sqrt 2) \in K, \tag{11}$

and

$a = a_1 + a_2\sqrt K \Longleftrightarrow -a = -a_1 - a_2\sqrt 2 \in K; \tag{12}$

(11) and (12) affirm that $K$ contains an additive identity element and that every element of $K$ is possessed of an additive inverse, also in $K$. Furthermore,

$1 = 1 + 0\sqrt 2 \in K, \tag{13}$

i,e,, $K$ contains a multipliciative identity as well.

So $(K, +)$ is an abelian group, also closed under multiplication, with multiplicative identity $1$. Commutativity, associativity, and distributivity of these operations hold in $K$ by virtue of the fact they are inherited from $\Bbb R$.

$K \subsetneq \Bbb R \tag{14}$

is a sub-ring. $K$ is thus itself a commutative unital ring, and we only need show every $a \in K$ has a multiplicative inverse to prove $K$ is in fact a field. To this end we observe that for any

$0 \ne a = a_1 + a_2\sqrt 2 \in K, \tag{15}$

$a_1^2 - 2a_2^2 \ne 0; \tag{16}$

otherwise

$a_1^2 = 2a_2^2, \tag{17}$

whence

$2 = \dfrac{a_1^2}{a_2^2}; \tag{18}$

but this is impossible since it yields

$\sqrt 2 = \dfrac{\vert a_1 \vert}{\vert a_2 \vert}, \tag{19}$

which asserts $\sqrt 2 \in \Bbb Q$; therefore (16) binds and we may write

$(a_1 + a_2\sqrt 2)\dfrac{a_1 - a_2\sqrt 2}{a_1^2 - 2a_2^2} = \dfrac{a_1^2 - 2a_2^2}{a_1^2 - 2a_2^2} = 1, \tag{20}$

so that

$(a_1 + a_2\sqrt 2)^{-1} = \dfrac{a_1 - a_2\sqrt 2}{a_1^2 - 2a_2^2}; \tag{21}$

we now may conclude that $K$ is indeed a field, since $(K, \cdot)$ forms an abelian group.

Robert Lewis
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    +1 There seems to be a typo in equation 11, and you reference equations (10) and (11) where I think you mean (11) and (12). – Servaes Jul 27 '19 at 17:57
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    @Servaes: Corrected. Thank you for pointing that out. And for your endorsement. Cheers! – Robert Lewis Jul 27 '19 at 18:00
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    Is it possible to show that $a^{-1}=\frac{1}{a_1+a_2\cdot \sqrt{2}} \overset{!}\in K$ – Analysis Jul 27 '19 at 18:28
  • Maybe like that: $$\frac{1}{a_1+a_2\cdot \sqrt{2}}\cdot \frac{a_1-a_2\cdot \sqrt{2}}{a_1-a_2\cdot \sqrt{2}}=\frac{a_1-a_2\cdot \sqrt{2}}{a_1^2-2a_2^2}=\underbrace{\frac{a_1}{a_1^2-2a_2^2}}{\in \mathbb{Q}}-\underbrace{\frac{a_2}{a_1^2-2a_2^2}}{\in \mathbb{Q}}\cdot \sqrt{2}$$ – Analysis Jul 27 '19 at 18:42
  • @ParabolicAlcoholic: Affirmative. Compare equations (20)-(21) in my answer. – Robert Lewis Jul 27 '19 at 18:46
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Here is a fast way to prove all non-zero elements of $K$ have multiplicative inverses in $K$, without computations:

Observe $K$ is a two-dimensional $\mathbf Q$-vector space and for any non-zero $x\in K$, consider the map \begin{align} m_x:K& \longrightarrow K,\\ y&\longmapsto xy. \end{align} This is a $\mathbf Q$-linear map, and as we are in a subring of the field $\mathbf C$, this linear map is injective. However, in a finite dimensional vector space, injectivity of an endomorphism is equivalent to its surjectivity. Thus $1$ is attained, i.e. there exists $y\in K$ such that $xy=1$ thereby proving the existence of multiplicative inverses.

Bernard
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Think about $\mathbb{C}$ here.

You note a complex number $z\in\mathbb{C}$ as $z=a+ib$.

An element $k\in K$ can be written as $k=x+\sqrt{2}y$

So to verify, that it is closed under addition, would go like this:

Take $k,k'\in K$. Then $k=x+\sqrt{2}y$ and $k'=x'+\sqrt{2}y'$.

Now $k+k'=x+\sqrt{2}y+x'+\sqrt{2}y'=(x+x')+\sqrt{2}(y+y')$. So it is of the form which is required to be an element of $K$. Note that $x+x'$ and $y+y'$ are elements of $\mathbb{Q}$.

Similar you can show, that $k\cdot k'$ is an element of $K$.

The only harder part is to show, that $k\neq 0$ (so $x,y\neq 0$) has an inverse $k^{-1}$.

But if you make yourself clear what is to show, I think you can do it.

This might be worth looking at:

Show that $\mathbb{Q}[\sqrt{2}] = \{a + b\sqrt{2}\}$ is a vector space over $\mathbb{Q}.$

Cornman
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  • Thanks, Cornman! Now I know how to start. – Analysis Jul 27 '19 at 16:55
  • You are welcome. Feel free to ask, if something is unclear. – Cornman Jul 27 '19 at 16:56
  • So for $kk'=(x\cdot x'+2y\cdot y')+\sqrt{2}(x\cdot y'+x'\cdot y) \in K$. Furthermore we have $0=0\cdot \sqrt{2}\cdot 0\in K$ and $1=1\cdot \sqrt{2}\cdot 0\in K$. Lastly,$-k=-x-\sqrt{2}y\in K$.... but I am having trouble with $k^{-1}$. – Analysis Jul 27 '19 at 17:08
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Well, first you can view the elements $x+\sqrt 2y$ of ${\Bbb Q}(\sqrt 2)$ as formal sums and prove that this set forms a field.

Second you can show that this field is isomorphic to the quotient ring ${\Bbb Q}[x]/\langle x^2-2\rangle$.

Bernard
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Wuestenfux
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