Is $f(\sqrt 2)$ in $\Bbb Q[\sqrt 2]$ where $f\in\Bbb Q[x]$?

Question

Prove that $\mathbb{Q}[x]/\langle x^2-2\rangle$ is ring isomorphic to $$\mathbb{Q}[\sqrt 2]=\{a+b\sqrt 2\mid a,b\in\mathbb{Q}\}.$$

Attempt: I think I have to apply first isomorphism theorem here and I will consider the function $\phi: f(x)\rightarrow f(\sqrt 2)$ then the $\ker\phi$ will be $\langle x^2- 2\rangle$. But I am confused if am I choosing the right function and is $f(\sqrt 2)$ is in $\mathbb{Q}[\sqrt2]$?

Answer 1

Define a map $\psi:\Bbb{Q}(\sqrt{2})\to \Bbb{Q}[x]/<x^2-2>$ by $\psi(a+b\sqrt{2})=[a+bx]$ for any $a,b\in\Bbb{Q}$.

$\psi$ is homomorphism:

$\psi((a+b\sqrt{2})+(c+d\sqrt{2}))=\psi((a+c)+(b+d)\sqrt{2}))=[(a+c)+(b+d)x]=[a+bx]+[c+dx]=\psi(a+b\sqrt{2})+\psi(c+d\sqrt{2}$.

$\psi((a+b\sqrt{2})(c+d\sqrt{2})=\psi((ac+2bd)+(ad+bc)\sqrt{2}))=[(ac+2bd)+(ad+bc)x]$. and, $\psi((a+b\sqrt{2})\psi(c+d\sqrt{2})=[a+bx][c+dx]=[ac+adx+bcx+bdx^2]=[ac+(ad+bc)x+bdx^2]=[(ac+2bd)+(ad+bc)x]$.

Therefore, $\psi$ is homomorphism.

$\psi$ is injective:

Let, $\psi(a + b\sqrt{2}) = \psi(c + d\sqrt{2})$, then $[a + bx] = [c + dx]$. Since there is a unique polynomial of degree $1$ or less for each congruence class $\operatorname{mod} p(x)$. $\Rightarrow$ $a + bx = c + dx$ $\Rightarrow$ $ a = c$ and $b = d$. Hence $\psi$ is injective.

$\psi$ is surjective:

Let $y\in \Bbb{Q}[x]/<x^2-2>$ then $y=[l+mx]$ for some $l,m\in \Bbb{Q}$. Therefore $\psi(l+m\sqrt{2})=y=[l+mx]$.

$\Rightarrow$ $\psi$ is surjective. And hence $\psi$ is isomorphism. Thus $\Bbb{Q}(\sqrt{2})\cong \Bbb{Q}[x]/<x^2-2>$.

Alternatively, Note that $\Bbb{Q}[x]/<x^2-2>$ is a field because $p(x)=x^2-2$ is irreducible over $\Bbb{Q}$ therefore $<p(x)>$ is maximal ideal. and $\Bbb{Q}(\sqrt{2})$ is the smallest field containing both $\sqrt{2}$ and $\Bbb{Q}$. Define a map $\phi:\Bbb{Q}[x]/<x^2-2>\to \Bbb{Q}(\sqrt{2})$ such that $\phi(a(x)+<x^2-2>)=a(\sqrt{2})$. $\phi$ is injective ring homomorphism since $\phi$ is not the zero map. The way map defined it is clear that $\phi$ is surjective. It is easy to see that. $\Bbb{Q}(\sqrt{2})=\Bbb{Q}[\sqrt{2}]$.

Answer 2

A way to tackle this type of problem uses the Euclidean algorithm. You could do the following:

Let $f(x) \in \mathbb{Q}[x]$. We now divide this by $x^2 - 2 \in \mathbb{Q}[x]$, obtaining: $f(x) = q(x) (x^2 - 2) + (ax + b)$, since the degree of the remainder is smaller than the degree of de divisor.

Apply the equivalence class function to get that, in the quotient ring, $\overline{f(x)} = \overline{q(x) (x^2 - 2) + (ax + b)}$, and, since the equivalence class of "$(x^2 - 2)$" is "$0$" and the quotient is a ring (because we divided by an ideal), we get $\overline{f(x)} = \overline{(ax + b)}$.

Now, all you have to do is show that $\overline{x}$ satisfies the equation $x^2 - 2 = 0$, and the isomorphism follows trivially.

Answer 3

Alternatively, consider the ring homomorphism $\varphi: \mathbb Q[x] \to \mathbb R$ induced by $x \mapsto \sqrt 2$. Then prove:

• the image of $\varphi$ is $\mathbb{Q}[\sqrt 2]$ – by definition

• $\mathbb{Q}[\sqrt 2]=\{a+b\sqrt 2\mid a,b\in\mathbb{Q}\}$ – even powers of $\sqrt2$ reduce to integers and odd powers reduce to an integer times $\sqrt2$; also follows from polynomial division.

• $\ker\varphi = \langle x^2- 2\rangle$ – follows from polynomial division.