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

I understand that posting questions without showing your work is looked down upon, however, speaking freely, I don’t know how to even begin the demonstration. I know how to demonstrate something is a vector space, I don’t know how to operate with $\mathbb{Q}[\sqrt{2}]$.

J. De Ro
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Sigma
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  • Which words and symbols do you understand? It's very hard for us to help you if we don't have any idea where you're getting stuck when you try to start. – jgon Apr 18 '19 at 17:52
  • Everything except how to deal with Q[sqrt{2}]. By the way I’m on my phone so formatting mathjax takes too long. – Sigma Apr 18 '19 at 17:55
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    The operations are the usual operations in $\mathbb{R}$. You know how to add real numbers and how to multiply a real number by a rational scalar. So show that $\mathbb{Q}[\sqrt{2}]$ is closed under these operations. – Mark Apr 18 '19 at 17:59
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    The easiest way is to identify $\mathbb{Q}[\sqrt{2}]$ as subspace of $\mathbb{R}$ as $\mathbb{Q}$ vector space. $\mathbb{Q}[\sqrt{2}]$ contains all numbers of the form $a+b\sqrt{2}$ with $a,b\in\mathbb{Q}$. Can you show that $0$ is of this form? Can you show that the sum of two expressions of this form has the same form again? Can you show that the product of such an expression with a rational has the same form again? If you manage to show these, you're done. – Thorgott Apr 18 '19 at 18:00

2 Answers2

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An $F$-vectorspace is an abelian group equipped with an $F$-action i.e. it is a set $V$ with a binary operation $+$ such that $(V,+)$ is an abelian group and another operation $\cdot : F \times V \to V$ satisfying field action axioms. I am sure you've seen the definition of a vectorspace before, but I believe it is helpful to go back to basics. Does this set

$$\mathbb{Q} = \{a+b\sqrt{2} : a,b \in \mathbb{Q}\}$$

satisfy the axioms of a $\mathbb{Q}$-vectorspace? What is addition? What is scalar multiplication?

Xabu
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This might get you started:

We have a field $\mathbb{Q}$ with the usual interpretation of $\oplus$ and $\odot$. I write $\oplus$ and $\odot$, which might look confusing, but I think it helps, because we actually want to seperate the usual $+$ and $\cdot$ you know from $(\mathbb{Q},+,\cdot)$ here. But, as it turns out, they are just the same.

Why is it still helpful? Because we want to know, where the structure comes from (laws of a associativity, commutativity, distributivity) and you already know, that these hold for the usual $+$ and $\cdot$.

First of all we have to show, that $\oplus:\mathbb{Q}[\sqrt{2}]\times\mathbb{Q}[\sqrt{2}]\to\mathbb{Q}[\sqrt{2}]$, $(x,y)\mapsto x+y$ is 'closed' under addition. So when you add two elements from $\mathbb{Q}[\sqrt{2}]$ then the sum is an element of $\mathbb{Q}[\sqrt{2}]$.

Note again, that $+$ is the sign you already know by heart.

So let $x,y\in\mathbb{Q}[\sqrt{2}]$. By definition of this set, it is $x=a+\sqrt{2}b$ and $y=c+\sqrt{2}d$, with $a,b,c,d\in\mathbb{Q}$.

Now we have to show, that $x+y\in\mathbb{Q}[\sqrt{2}]$. That means it has to be $x+y=a'+\sqrt{2}b'$ for some $a',b'\in\mathbb{Q}$.

And finding them is a simple task, since

$\underbrace{\oplus(x,y)}_{\text{Again, this ist just notation}}=\underbrace{(a+\sqrt{2}b)+(c+\sqrt{2}d)}_{\text{by definition of the binary operation $\oplus$}}=\underbrace{(a+c)}_{=a'}+\sqrt{2}\underbrace{(b+d)}_{=b'}$

So indeed we have $x+y\in\mathbb{Q}[\sqrt{2}]$.

Note that we use for the last equality, that $+$ and $\cdot$ (where I used the common notation of $x\cdot y=xy$) already has these properties of associativity, commutativity and distributivity.

There are a bunch of axioms to show, all are pretty much really simple, if you just try. Let me show one more:

We have to give a neutral element, with regards to $+$, for $(\mathbb{Q}[\sqrt{2}],\oplus,\odot)$ to be a vectorspace.

This is given by $0+0\sqrt{2}=0$

Because it is $\oplus(x,0)=(a+\sqrt{2}b)+(0+\sqrt{2}0)=(a+0)+\sqrt{2}(b+0)=a+\sqrt{2}b=x$

And $\oplus(0,x)=(0+\sqrt{2}0)+(a+\sqrt{2}b)=(0+a)+\sqrt{2}(0+b)=a+\sqrt{2}b=x$

Note, that $0$ already has this property in the field $(\mathbb{Q}, +,\cdot)$.

The other axioms can be shown this way. I hope this little guide helps.

Cornman
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  • It helped. I understand now that $(a + b\sqrt{2})$ is to be treated as a vector. The rest follows easily. Thanks. – Sigma Apr 18 '19 at 18:42
  • Otherwise it might help, when you study how you show, that $\mathbb{C}$ is a field. Because there you write the elements as (a,b) where a is the real part and b is the imaginary part. Here you can write this as (a,b) too, as you said "treat it like a vector". Where a is the rational part and b is the $\sqrt{2}$-part. But it is simpler, since $\mathbb{C}$ has an odd multiplication, which $\mathbb{Q}[\sqrt{2}]$ has not. – Cornman Apr 18 '19 at 18:46
  • I have shown that C is a field. – Sigma Apr 18 '19 at 18:51