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If I have an ideal $(x-1)$ for the ring $\Bbb R[x]$, how do I think of the quotient ring $\Bbb R[x]/(x-1)$? I have all polynomials with:

$$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0 {\pmod {x-1}}$$ How do I think of the Simple ring generated by this? What is an arbitrary element in this quotient ring?

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  • $\begingroup$ It looks like the polynomial evaluated at $x=1$. $\endgroup$ – jgon May 22 '15 at 6:56
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When you quotient by $x-1$ you are saying that $x-1 \equiv 0$ (or $x \equiv 1)$. With this in mind \begin{align*} a_nx^n+a_{n-1}x^{n-1}+ \dotsb +a_1x+a_0 & \equiv a_n(1)^n+a_{n-1}(1)^{n-1}+ \dotsb + a_1(1)+a_0 \pmod{x-1}\\ & \equiv a_n+a_{n-1}+ \dotsb + a_1+a_0 \pmod{x-1}\\ \end{align*} Thus every polynomial is equivalent to its value at $x=1$. Thus $$\mathbb{R}[x]/\langle x-1\rangle \cong \mathbb{R}$$ and the cosets would look like $$\mathbb{R}[x]/\langle x-1\rangle = \{r+\langle x-1\rangle \, | \, r \in \mathbb{R}\}.$$

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