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?
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Isomorphic as what? – Rasmus Feb 25 '13 at 21:03
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as rings (subrings of $R$) – Federica Maggioni Feb 25 '13 at 21:15
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As rings or as subrings of $R$? Sorry, I'm not trying to pick on you. That makes a difference. – Rasmus Feb 25 '13 at 21:30
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let me think...i'm interested to ideals of $R$. So i think: suppose i know $\frac{R}{\mathfrak{m}}$ and $\frac{R}{\mathfrak{m'}}$ are isomorphic as rings. Can i say that $\mathfrak{m}$ and $\mathfrak{m'}$ is the "same" ideal of $R$? So i'm asking if $\mathfrak{m}$ and $\mathfrak{m'}$ are isomorphic as $R$-modules – Federica Maggioni Feb 25 '13 at 21:36
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I don't think being isomorphic as $R$-modules is the same as being isomorphic as ideals in $R$ either because you are forgetting the multiplication. – Rasmus Feb 26 '13 at 07:29
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the multiplication as $R$-modules is the same as the multiplication as subrings of $R$ in this case. Isn't it? – Federica Maggioni Feb 26 '13 at 07:50
3 Answers
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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.
Tom Oldfield
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Nice example. Is it obvious that $C[1/2,1]$ is isomorphic to $C[0,1]$? I don't have pen/paper, but is the shifting/scaling of the function's argument a homomorphism? – Jonathan Feb 25 '13 at 21:16
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@Jonathan Thank you very much! Yes, that's exactly the one I was thinking of, when you have some pen and paper I suggest you quickly write it out! – Tom Oldfield Feb 25 '13 at 21:24
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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.
Zev Chonoles
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Let $\rm\:R = \Bbb Q[x_1,x_2,x_3,\ldots].\:$ Then $\rm\: R\,\cong\, R/(x_1\!)\,\cong\, R/(x_1,x_2)\cong R/(x_1,x_2,x_3)\,\cong\, \cdots$
Math Gems
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