The Ring of Dual Numbers

Detailed Solution

By definition, $\varepsilon\neq0$ but $\varepsilon^2=0$, so $\varepsilon$ is a nonzero nilpotent element of $R[\varepsilon]$. This immediately rules out option A: a field can have no nonzero nilpotents, since if $x\neq0$ were nilpotent with $x^n=0$, then multiplying by $x^{-1}$ repeatedly would force $x=0$, a contradiction. It also rules out option C, since $\varepsilon\cdot\varepsilon=0$ with $\varepsilon\neq0$ is precisely a zero-divisor relation. For option D, check whether the proposed map $\phi(a+b\varepsilon)=(a,a+b)$ is a ring homomorphism. Compute the product in $R[\varepsilon]$: $$ (a+b\varepsilon)(c+d\varepsilon)=ac+(ad+bc)\varepsilon, $$ using $\varepsilon^2=0$. So $$ \phi\big((a+b\varepsilon)(c+d\varepsilon)\big)=(ac,\;ac+ad+bc). $$ But the componentwise product in $R\times R$ gives $$ \phi(a+b\varepsilon)\cdot\phi(c+d\varepsilon)=(a,a+b)\cdot(c,c+d)=(ac,\;(a+b)(c+d))=(ac,\;ac+ad+bc+bd). $$ These two results differ by the extra term $bd$, so $\phi$ is not multiplicative and hence not a ring isomorphism. In fact $R[\varepsilon]$ cannot be isomorphic to $R\times R$ at all, since $R\times R$ has no nonzero nilpotent elements while $R[\varepsilon]$ does (namely $\varepsilon$). Thus the correct option is $\boxed{\mathrm{B}}$.