Looking for an example of quotient ring isomorphism.

Question

In a homework problem, I need to find an example of quotient ring isomorphism $\frac{R}{I}\cong\frac{R}{J}$ such that $I\neq J$. I think that it is a common issue but I am not pretty sure about it...

I have been reading about this issue and I have found that if I take any polynomial ring $R=K[X]$ for some field $K$, if I consider the ideal $I=(X)$ and the ideal $J=(X-1)$ then $\frac{R}{I}\cong\frac{R}{J}\cong K$.

It is true? Could you give me a specific example with a little clarification or comment?

Answer 1

Yes note that since K is a field K[x] is an Euclidean domain. Now map you want to consider is sending polynomial to its remainder. (This is almost an answer I won't elaborate more as this is a homework.)

Answer 2

For $a\in \mathbb{R}$, fuction: $$\phi_a:\mathbb{R}[x]\to \mathbb{R}:\phi_a(p(x)):=p(a)$$ is a ring homeomorphism: $$\phi_a(p(x)+q(x))=p(a)+q(a)=\phi_a(p(x))+\phi_a(q(x))$$ $$\phi_a(p(x)q(x))=p(a)q(a)=\phi_a(p(x))\phi_a(q(x)).$$ Also:

$$\mathrm{Ker}\,\phi_a=\{p(x)\in\mathbb{R}[x]:\phi_a(p(x))=p(a)=0\}= \{p(x)\in\mathbb{R}[x]:x-a | p(x)\}=<x-a>$$

and $$\mathrm{Im}\,\phi_a=\{\phi_a(p(x))=p(a): p(x)\in\mathbb{R}[x]\}=\mathbb{R}.$$ So function: $$\mu_a:\frac{\mathbb{R}[x]}{<x-a>}\to \mathbb{R}:\mu_a\big(p(x)+<x>\big)=p(a)$$ is a ring isomorphism. So $$\frac{\mathbb{R}[x]}{<x-a>} \cong \frac{\mathbb{R}[x]}{<x-b>}\cong \mathbb{R},$$ while $<x-a>\neq <x-b>$ for $a\neq b.$