rings and ideals isomorphisms Let $R$ be a ring and $\mathfrak{m},\mathfrak{m'}$ two ideals of $R$.
Suppose that $\frac{R}{\mathfrak{m}}$ and $\frac{R}{\mathfrak{m'}}$ are isomorphic. Can i san say that $\mathfrak{m}$ and $\mathfrak{m'}$ are isomorphic too?
 A: Unfortunately not!
For a particularly interesting example, consider $R = C[0,1]$, the ring of continuous functions from $[0,1] \rightarrow R$. Take the ideal $I = \{f \in R : f(x) = 0 \space \forall x \in [0,\frac12]\}$. The quotient $R/I$ is then isomorphic to $C[\frac12, 1]$, which is in turn isomorphic to $R$ itself! 
Certainly then $R/I \cong R/\{0\}$, but $I \not\cong{0}$. This is also shows that in a general ring, information about the quotient may contain very little information about the ideal being quotiented out.
A: No; for example, let $R=\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/4\mathbb{Z}$, let $$\mathfrak{m}=\mathbb{Z}/2\mathbb{Z}\times2\mathbb{Z}/4\mathbb{Z}=\{(\overline{0},\overline{0}),(\overline{1},\overline{0}),(\overline{0},\overline{2}),(\overline{1},\overline{2})\}$$
and
$$\mathfrak{m}'=0\times\mathbb{Z}/4\mathbb{Z}=\{(\overline{0},\overline{0}),(\overline{0},\overline{1}),(\overline{0},\overline{2}),(\overline{0},\overline{3})\}.$$ Then
$$R/\mathfrak{m}\cong R/\mathfrak{m}'\cong\mathbb{Z}/2\mathbb{Z}$$
but $\mathfrak{m}$ and $\mathfrak{m}'$ are not isomorphic as abelian groups, and hence certainly not isomorphic as $R$-modules.
A: Let $\rm\:R = \Bbb Q[x_1,x_2,x_3,\ldots].\:$ Then $\rm\: R\,\cong\, R/(x_1\!)\,\cong\, R/(x_1,x_2)\cong R/(x_1,x_2,x_3)\,\cong\, \cdots$
