showing $Q[\sqrt 2] = Q(\sqrt 2)$

Question

The question came in my exam. $Q[\sqrt 2] = \{ a + b \sqrt2 \;| a,b \in Q\}$ and $Q(\sqrt 2)$ is minimal subfield of it's extension containing $Q$ and $\sqrt 2$.

(In my book) It calls $F(a)$ adjoining $a$ to $F$ where $a$ is element of finite extension $E$ of F. I am not sure about my question (probably not nice since my question bank contains some error), any hints (and correction) will be appreciable. Thanks!!

Answer 1

Hint :

show that $a+b\sqrt{2}\in \mathbb{Q}[\sqrt{2}]$ is invertible.

Qn : Is it sufficient?

Answer 2

Before all is clear that $\mathbb Q[\sqrt2]\subseteq\mathbb Q(\sqrt2)$. Let's see the opposite inclusion.

Consider the evaluation morphism $ev_{\sqrt{2}}:\mathbb{Q}[X]\longrightarrow\mathbb{Q}[\sqrt2]$. It fixes costant elements (i.e. $q\mapsto q$ $\forall q\in\mathbb Q$) and $X\mapsto\sqrt2$. In such a way, being $ev_\sqrt 2$ an homomorphism of rings, it's completely determinated.

Now you see it's clearly surjective and $\ker(ev_\sqrt2)=(X^2-2)$. Hence by first homomorphism theorem you immediately have that $\mathbb Q[\sqrt2]\simeq \frac{\mathbb Q[X]}{(X^2-2)}$. But the polynomial $X^2-2$ is irreducible in $\mathbb Q[X]$, hence the ideal $(X^2-2)$ is maximal, so the quotient $\frac{\mathbb Q[X]}{(X^2-2)}$ is a field and therefore $\mathbb Q[\sqrt2]$ is a field too. Hence now you have two fields, $\mathbb Q(\sqrt2)$ and $\mathbb Q[\sqrt2]$, containig both $\mathbb Q$ and $\sqrt2$. But for definition $\mathbb Q(\sqrt2)$ is the smallest of such fields. So $\mathbb Q(\sqrt2)\subseteq\mathbb Q[\sqrt2]$, as wanted.

So you can conclude that $\mathbb Q(\sqrt2)=\mathbb Q[\sqrt2]$.

Answer 3

As I said $\mathbb Q(\sqrt2)$ is the smallest field contianing $\mathbb Q$ and $\sqrt2$. This by definition! You have that $\mathbb Q(\sqrt2):=Frac(\mathbb Q[\sqrt2])$... but this is valid for every domain $R$; you can always consider $Frac(R)$ and by construction you can see there is a natural immersion $R\hookrightarrow Frac(R)$ given by $a\longmapsto\frac{a}{1_{R}}$ (keep in mind the construction you see in gettin' rational numbers from integers: you have an immersion $\mathbb Z\hookrightarrow\mathbb Q$ given by $h\longmapsto\frac{h}{1}$).

If you want you can think at $\mathbb Q(\sqrt2)$ as the intersection of all subfields of $\mathbb R$ containing $\mathbb Q$ and $\sqrt2$, i.e. $\mathbb Q(\sqrt2):=\bigcap_{F\leq\mathbb R \;s.t.\; \mathbb Q \subseteq F \;and\;\sqrt2\in F}F$, but this isn't useful for our purpose.

But if you want to see explicitly that $\mathbb Q[\sqrt2]\subseteq\mathbb Q(\sqrt2)$, having in mind that $\mathbb Q(\sqrt2)$ is by def. a field (the smallest, but now it's not relevant) containing both $\mathbb Q$ and $\sqrt2$, it should be clear that $q\in\mathbb Q(\sqrt2)\;\;\forall q\in\mathbb Q$, $\sqrt2\in\mathbb Q(\sqrt2)\Longrightarrow q\sqrt2\in\mathbb Q(\sqrt2)$ hence $q+p\sqrt2\in\mathbb Q(\sqrt2)$ for every $p,q\in\mathbb Q$. But these ones are all and only the elements of $\mathbb Q[\sqrt2]$, so we're saying exactly that $\mathbb Q[\sqrt2]=\{q+p\sqrt2\;\ :\;q,p\in\mathbb Q\}\subseteq \mathbb Q(\sqrt2)$.

Pay attention! Before you prove that $\mathbb Q[\sqrt2]=\mathbb Q(\sqrt2)$ you can't know $\mathbb Q[\sqrt2]$ is a field... but you know this is a ring. Moreover $\mathbb Q(\sqrt2)$ is a field by definition. But every field is a ring, so what you have is an inclusion of rings!