I guess no, because $x^3$ is not divisible by $x^2$, but I am not sure whether it is right to think in this way or not.
Remarks:
R[x] denotes the set of all polynomials with coefficients in R. $R[x]/(x^2)$ is the quotient ring.
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1is $R$ a ring ? – IntegrateThis Apr 15 '19 at 06:09
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Yes, they are both rings. – bbb Apr 15 '19 at 06:26
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Either I am missing something or this is true by recognizing they both contain $0$ as the only root. – IAmNoOne Apr 15 '19 at 06:31
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Isomorphic as rings or as R-modules? – Frieder Jäckel Apr 15 '19 at 06:33
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Isomorphic as rings. But i am also interested to know what are the differences between isomorphic as rings or as R-modules? – bbb Apr 15 '19 at 06:36
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1If $R$ is a commutative unitary ring, then $R[X]/(X^2)$ is a free $R$-module of rank $2$, while $R[X]/(X^3)$ is a free $R$-module of rank $3$. Therefore they are not isomorphic as $R$-modules. – user26857 Apr 15 '19 at 21:42
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If $R=\Bbb Q[x_1,x_2,\dots;y_1,y_2,\dots]/(x_1^2,x_2^2,\dots;y_1^3,y_2^3,\dots)$, I believe that $R[x]/(x^2)\cong R[x]/(x^3)$ (and indeed both quotients are isomorphic to $R$ itself).
Greg Martin
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1Sorry for not stating clearly. $R[x]$ denotes the set of all polynomials with coefficients in R. – bbb Apr 15 '19 at 06:54
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Am I correct in saying that the map defined by sending $x$ to $x_1$ and $x_i$ to $x_{i+1}$ is an isomorphism of rings from $R[x]/(x^2)$ to $R$? – abhi01nat Apr 15 '19 at 17:00
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