This is my (informal) understanding of a quotient ring. I understand that this is very flimsy, but I hope you can get the main idea.

You have some ring $R$ and you want to quotient out an ideal $I$. Draw out a 'number line' listing all the elements, starting from the zero element. Identify the smallest element of $I$. Take only the elements from the zero element up to that smallest element of the ideal (ideal generator), delete everything else. You now have a 'string' of elements. Connect the ends of the string so the ideal generator $I$ coincides with the zero element. This is your quotient 'ring'.

So going by this, the quotient ring should contain all the elements 'smaller than' the ideal generator. So for example, if the ideal generator is $x^2+1$, then the ring contains only linear factors, hence 'smaller'.

This is the part where I am confused. My understanding is that if we have $$R[x]/(x^2+1) \cong ax+b$$

for $a,b \in R$.

But this excellent answer here says that

$$\mathbb{Q}[x]/(x^2-2) \cong \mathbb{Q}[\sqrt{2}]$$

Why are we finding the roots of the quotient and plugging them into the quotient form (linear factor)?

  • 1
    $\begingroup$ The statement $R[X]/(X^2+1)\simeq aX+b$ is meaningless. $\endgroup$ – Pedro Tamaroff Jun 13 '16 at 3:24
  • $\begingroup$ here, $R_1 \cong R_2$ means that $R_1,R_2$ are two different rings but with a ring isomorphism between the two, that is, there exists $\phi : R_1 \to R_2$ such that $\phi(xy+vw) = \phi(x)\phi(y)+\phi(v)\phi(w)$, and the same for $\phi^{-1}$ $\endgroup$ – reuns Jun 13 '16 at 3:26
  • $\begingroup$ and $R[x] / (x^2+1)$ is the set of cosets $P+ (x^2+1), P \in R[x]$, for each of those we can choose a representative $ax+b$, with the same $+$ as in $R[x]$, but with a different $\times$ (can you write the multiplication $(ax+b)(cx+d)$ in $R[x] / (x^2+1)$ ?) $\endgroup$ – reuns Jun 13 '16 at 3:28

When you quotient out by $x^2+1$ you send every multiple of $x^2+1$ to zero. This is the same as making $x^2=-1$ which is the same as making $x=i$ so $R[x]$ is now $R[i]$ It's probably the wrong way to think about it and I'm sure I will be scolded for it, but when you just want to know what the damn ring is that's the easiest way to think about it.


To answer the last part of your question, define a ring homomorphism $\varphi : \mathbb{Q}[x] \to \mathbb{Q}[\sqrt{2}]$ by $\phi(f) = f(\sqrt{2})$. It is easy to see that $\varphi$ is surjective. Since $m(x) = x^2 - 2$ is the minimal polynomial of $\sqrt{2}$, $m$ must divide any polynomial $g \in \mathbb{Q}[x]$ with $g(\sqrt{2}) = 0$. Therefore $\ker\varphi = (x^2 - 2)$. By the first isomorphism theorem, $$\frac{\mathbb{Q}[x]}{(x^2 - 2)} \cong \mathbb{Q}[\sqrt{2}].$$


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