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!!


Hint :

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

Qn : Is it sufficient?

  • $\begingroup$ the OP is my other account for asking stupid questions, for some reason I can't log in. I can' show that it is invertible, multiplying by conjugate, but how does it show?? $\endgroup$ – Santosh Linkha Apr 5 '14 at 11:32
  • $\begingroup$ @SantoshLinkha are you happy that you've then shown $\mathbb Q [\sqrt2]$ is a field? Now $\mathbb Q (\sqrt2)$ is the minimal extension of $\mathbb Q$ containing $\sqrt 2$... $\endgroup$ – ah11950 Apr 5 '14 at 11:34
  • $\begingroup$ @SantoshLinkha : Does above comment clear your confusion? $\endgroup$ – user87543 Apr 5 '14 at 11:35
  • $\begingroup$ @ah11950 yes I understand. I'll have to wait 5 hrs before I can log in to that account. and thank you!! $\endgroup$ – Santosh Linkha Apr 5 '14 at 11:36
  • $\begingroup$ @PraphullaKoushik I got it now :-) $\endgroup$ – Magdiragdag Apr 6 '14 at 7:14

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 $\mathbb Q(\sqrt2)=\mathbb Q[\sqrt2]$.

  • $\begingroup$ Hi Joe. Forgive my possible ignorance. But the statement " Before all is clear que $ \mathbb{Q}[\sqrt{2}] \subseteq \mathbb{Q}(\sqrt{2})$ " is not at all clear. $\endgroup$ – MathOverview Apr 5 '14 at 12:48
  • $\begingroup$ $\mathbb Q [\sqrt 2]$ is the set of all $\mathbb Q$-linear combinations of $1, \sqrt 2$. Any field extension of $\mathbb Q$ containing $\sqrt 2$ must contain all such linear combinations (as it's a field, it's certainly closed under multiplication and addition). $\endgroup$ – ah11950 Apr 5 '14 at 13:46

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!


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