Showing that every element in $\mathbb Q$ $[x]$ / $(x^2-2)$ can be written as $a+b\sqrt2$

Question

I'm having trouble showing this. I already know that $\sqrt2$ is the root of $(x^2-2)$ in $\mathbb Q$ $[x]$ / $(x^2-2)$ but why does this mean that every element can be written uniquely in the form $(a+b\sqrt2)$?

I tried looking at the theorem saying that an element of an extended field can be written as $$a(x)=f(x)*q(x)+r(x) $$ such that $[a(x)]=[r(x)]$. because there seems to be similarities but i can't really make it fit.

Answer 1

If this is how the problem is really phrased, I take a slight issue with this. There is a canonical isomorphism of the ring $E=\mathbb{Q}[x]/(x^2-2)$ with the ring $F=\mathbb{Q}(\sqrt{2})$, the latter of which is literally all real numbers of the form $a+b\sqrt{2}$ for $a, b \in \mathbb{Q}$. But elements of $E$ are cosets of the form $f(x) + (x^2-2)$ where $f(x) \in \mathbb{Q}[x]$ and not real numbers like $a+b\sqrt{2}$.

Here's what they're getting at. Take an element $f(x) + (x^2-2)$ in $E$. Long divide $f(x)$ by $x^2-2$ to write $$ f(x) = (x^2-2)q(x)+r(x) $$ with $r(x)=0$ or $\deg r(x) \leq 1$. Then clearly the cosets $f(x) + (x^2-2)$ and $r(x) + (x^2-2)$ are equal since $f(x)-r(x) \in (x^2-2)$. But, $r(x)=a+bx$ for some uniquely determined $a, b \in \mathbb{Q}$, so every element $f(x) +(x^2-2)$ of $E$ has a unique expression as $a+bx +(x^2-2)$. So, linear (or less) representatives suffice.

In the above, $x$ plays the role of $\sqrt{2}$, giving the desired representation (or, rather, "interpretation"). This is what happens in the isomorphism mentioned in the first paragraph. Evaluation at $x=\sqrt{2}$ defines a ring homomorphism $\phi: \mathbb{Q}[x] \to \mathbb{Q}(\sqrt{2})$. This is clearly onto since $a+b\sqrt{2}$ is $\phi(a+bx)$. (Look familiar?) Certainly $x^2-2$ is in the kernel of $\phi$, and a long-division argument like the one above shows that $\ker(\phi)=(x^2-2)$. (I'm also using facts about ideals in polynomial rings over a field here.) The fundamental homomorphism theorem then shows that $\phi$ induces an isomorphism $\mathbb{Q}[x]/(x^2-2) \cong \mathbb{Q}(\sqrt{2})$, as claimed above. Under this map $\phi$, note that $x$ maps to $\sqrt{2}$. That's what's going on in this exercise.

Answer 2

Consider the ring homomorphism $\varphi:\mathbb{Q}[x]\to \mathbb{Q}(\sqrt{2}):x\mapsto\sqrt{2}$. What is the kernel of that homomorphism? If you work it out, you will see that the kernel is the ideal $(x^2-2)$, so that there is a natural isomorphism between $\mathbb{Q}[x]/(x^2-2)$ and $\mathbb{Q}(\sqrt{2})$. Every element in the latter ring can clearly be written as $a+b\sqrt{2}$, $a$, $b\in \mathbb{Q}$.

Answer 3

The wording of the question is not completely right, but it is true that $ \mathbb{Q}[x]/(x^2 - 2) \cong \mathbb{Q}(\sqrt{2}) $; in the former, it can be shown via an induction argument that every element of the former can be represented by a polynomial of $ \mathbb{Q}[x] $ of degree less than 2; in the latter, the elements are as you said, written of the form $ a + b \sqrt{2}, a, b \in \mathbb{Q} $. The idea behind it is, $ x^2 - 2 $ has no roots in $ \mathbb{Q} $, but if you quotient this polynomial out, it is now the zero coset. In particular, $ \overline{x} = x $ is a root since by our quotienting relation, $x^2 \cong 2 $. The quotienting process identifies $x^2 - 2 $ with 0, same as saying $x^2 $ is identified with 2. Using the lemma I mentioned, you can then find the natural isomorphism $ \mathbb{Q}[x]/(x^2 - 2) \cong \mathbb{Q}(\sqrt{2}), \overline{x} \mapsto \sqrt{2} $.