Is $\mathbb R[x]/\langle (x-a)^2 \rangle $ isomorphic with some known ring ( where $a$ is a real number ) ? In particular is $\mathbb R [x] / \langle (x-1)^2 \rangle$ isomorphic with some known ring ? How many ideals do such quotient rings have ? I can determine that $\mathbb R[x]/\langle(x-a)(x-b)\rangle \cong \mathbb R \times \mathbb R$ , if $a,b$ are distinct real numbers , but having trouble if they are same ; please help . Thanks in advance .

$\Bbb EDIT$ : I am looking for a isomorphic ring which will simplify the quotient structure , loosely speaking , which will not be a quotient ring ... Thanks


There is a ring isomorphism $\Bbb R[x] \to \Bbb R[x+a] = \Bbb R[x]$, determined by $x \mapsto x+a$, and under this map, the polynomial $(x-a)^2$ maps to $x^2$. Therefore, the ring $\Bbb R[x]/\langle(x-a)^2\rangle$ is isomorphic to $\Bbb R[x]/\langle x^2\rangle$. Its elements can be represented in the form $a+bx$, where $a,b\in\mathbb R$. This ring has nilpotent elements (e.g., $x$) and thus is not isomorphic as a ring to $\mathbb R\times \mathbb R$, which has no such elements.

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  • $\begingroup$ Yeah , that I know :) but that does not simplify the quotient structure ... $\endgroup$ – user217921 Mar 4 '15 at 14:13
  • $\begingroup$ What would simplifying the quotient structure mean? The ring does not decompose as a direct product, so I would say there's not really a way to simplify it further than this. Although, if you'd like you could represent it as a matrix ring, $R=\{\pmatrix{a & b \\ 0 & a} : a,b\in\Bbb R\}$. $\endgroup$ – Brent Kerby Mar 4 '15 at 14:17
  • $\begingroup$ Uh huh , ok , then could you please say the homomorphism which maps the elements of $\mathbb R[x]$ to the matrix , you have mentioned , surjectively , with kernel $\langle x^2 \rangle$ ? $\endgroup$ – user217921 Mar 4 '15 at 14:21
  • $\begingroup$ I think the diagonal entries are $P(0)$ and off-diagonal entries $0$ and $P'(0)$ , for $P(x) \in \mathbb R[x]$ , right ? $\endgroup$ – user217921 Mar 4 '15 at 14:25
  • $\begingroup$ Take $a+bx \mapsto \pmatrix{a & b \\ 0 & a}$. It's equivalent to the regular representation of $\mathbb{R}/\langle x^2\rangle$, i.e. the representation obtained by considering the linear transformations induced by elements of $\mathbb{R}/\langle x^2\rangle$ acting on $\mathbb{R}/\langle x^2\rangle$ by multiplication. $\endgroup$ – Brent Kerby Mar 4 '15 at 14:26

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