Setting Ideal to Zero The professor introduced factor groups of a ring in class today. In particular she provided the following example:
$$\mathbb{R}[x]/\langle x-1\rangle \cong \mathbb{R}$$
Where $\mathbb{R}[x]$ are polynomials with real coefficients. She went through a computation of manipulating cosets to show the cosets have the same structure as $\mathbb{R}$. Then she said we could see this quickly by setting $$x-1=0 \\ \implies x=1$$ Plugging in $1$ into all real polynomials and observing this will result in only real numbers gives the result. I can see that this works formally but the equality of $x-1=0$ is confusing. The LHS is a real polynomial. The RHS is a label we give to that cosets in the factor group. They are objects from different sets! Can I just set these two equal and solve for $x$?
 A: As far as I can understand, you're looking for something like this:
It is $\mathbb{R}[x]/\langle x-1\rangle =\{f(x)+\langle x-1\rangle : f(x)\in\mathbb{R}[x]\}$. By the Euclidean division algorithm, one has $\mathbb{R}[x]/\langle x-1\rangle =\{q(x)(x-1)+r(x)+\langle x-1\rangle : $ where $ r(x)=0$ or $\deg(r(x))=0\}=\{r(x)+\langle x-1\rangle :$ where $r(x)=0$ or $\deg(r(x))=0\}=\{r+\langle x-1\rangle : r\in\mathbb{R}\}$, since polynomials of degree $0$ are constant. Now, the natural mapping $r\mapsto r+\langle x-1\rangle $ is an isomorphism from $\mathbb{R}$ to $\mathbb{R}[x]/\langle x-1\rangle $ and that should do it. For an application of the above, let's look at the example you're providing:
It is $x+\langle x-1\rangle =r(x)+\langle x-1\rangle $ where $r(x)$ is the remainder after we divide $x$ with $x-1$ on $\mathbb{R}[x]$. So $r(x)=1$. By what was stated above, $x+\langle x-1\rangle =1+\langle x-1\rangle $.
P.S: Always keep in mind that $f(x)+\langle g(x)\rangle =h(x)+\langle g(x)\rangle $ if and only if $f(x)-h(x)\in \langle g(x)\rangle $ if and only if $g(x)|f(x)-h(x)$.
A: The canonical map $\;\mathbf R[x]/(x-1)\longrightarrow\mathbf R$ is but the map $p(x)\longmapsto p(1)$, since the Euclidean division of $p(x)$ by $x-1$ results in a polynomial $q(x)$ such that
$$p(x)=q(x)(x-1)+p(1).$$
