I'm currently studying ring and field theory and I'm a little confused about whats going on behind the scenes of Field extensions,as I missed some course work last semester.

say we want to adjoin $\sqrt{2}$ to the field of rationals $ \Bbb Q $

1.we know that $ \Bbb Q(\sqrt{2}):=\{a+b\sqrt{2}| a,b \in \Bbb Q \} $

2.we also know that $ \Bbb Q[x]/<x^2-2> \cong \Bbb Q(\sqrt{2})$

my question is how do we get the form of $\Bbb Q(\sqrt{2})$ to be $a+b\sqrt{2}$ from the relation given in 2. looking around stack exchange i have seen some people say that we can obtain the representatives of the coset through long division ? I don't understand what is meant by this , but this more mechanical approach to understanding would be preferable to a theorem heavy explanation. If anyone can give any insights it would be much appreciated.


2 Answers 2


The point is that in $(2)$, you have set the polynomial $x^2-2=0$. Let $f(x)\in \mathbb{Q}[x]$ be any polynomial. It may be written (uniquely) as $$ f(x) = a_0+a_1x+a_2x^2+\cdots +a_nx^n.$$

Note now that since $x^2 = 2$ (so $x^{2n} = 2^n$) in $\mathbb{Q}[x]/\langle x^2-2\rangle$, the image of $f$ in this quotient may be written as $f(x) = b_0+b_1x$, where $$b_0 = a_0+2a_2+\cdots 2^ka_{2k},$$ and similarly for $b_1$. Thus, every element in $\mathbb{Q}[x]/\langle x^2-2\rangle$ may be written uniquely as $a+bx$ where $x^2=2$. This gives an isomorphism $$\mathbb{Q}[x]/\langle x^2-2\rangle\cong \mathbb{Q}(\sqrt{2}).$$

  • $\begingroup$ ah okay. I think the thing that was throwing me was that $a,b \in \Bbb Q$ instead of being elements of $\Bbb Q[x]$ but you're saying that because of the relation $x^2=2$ we can term $b_0$ the sum of $a_0$ and all the even powers of x, while $b_1$ is the sum of all odd powers of x, which is why we pull out the x here, and then $b_0$and$b_1$ now both represent rationals as x^2 has transformed to two ? $\endgroup$
    – excalibirr
    Mar 21, 2018 at 22:42
  • $\begingroup$ Yes that is correct $\endgroup$ Mar 22, 2018 at 12:40
  • $\begingroup$ I really want to upvote your comment again. Best explanation I've seen for this. It's really helped me understand alot about fieldextensions :) $\endgroup$
    – excalibirr
    Mar 22, 2018 at 17:02
  • $\begingroup$ I am glad to hear that :) $\endgroup$ Mar 22, 2018 at 20:17

An element of the quotient ring from (2) is (the equivalence class of) a polynomial P(x).

Now instead of thinking in terms of cosets, you can think of a quotient as a new rule added to the set of rules of a game.

  • Initial set of rules: axioms of rings.
  • New rule: now on top of that $x^2-2=0$. You set to $0$ whatever lies in the ideal you quotient by.

This can be re-written as $x^2=2$, which is a strong incentive to use a new notation and write $\sqrt{2}$ instead of $x$.

With this insight, you can write down the details (by Euclidean division any polynomial $P$ can be written as $a+bx$ plus some factor of $x^2-2$, which is identified with $0$, and the map induced by $x\mapsto \sqrt{2}$ is an isomorphism between the relevant structures).


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