Why does the computation make sense?

Question

An example using Kronecker's Theorem

Just wondering why $\alpha^2+1=0$ in this computation?

I also would like to know what $\langle x^2+1 \rangle$ actually is in the image of the above example. Is it the principal ideal generated by $x^2+1$? ie $x^2+1$ added n times such that n is in $\mathbb{R}$?

Finally why is $(x^2+1)+ \langle x^2+1 \rangle$ zero?

Answer 1

The zero element in the quotient ring $\mathbb{R}[x]/\langle x^2+1\rangle$ is $0+\langle x^2+1\rangle=\langle x^2+1\rangle$, which the author simply writes as $0$. Now, for any polynomials $f,g\in \mathbb{R}[x]$, we have $$ f+\langle x^2+1\rangle = g + \langle x^2+1\rangle \iff f-g\in\langle x^2+1\rangle $$ Since $x^2+1=(x^2+1)-0\in \langle x^2+1\rangle$, we deduce that $x^2+1+\langle x^2+1\rangle=0+\langle x^2+1\rangle$.

Answer 2

Back-to-front...

$x^2+1 \in \langle x^2 + 1 \rangle$, so $(x^2 + 1) + \langle x^2 + 1 \rangle = \langle x^2 + 1 \rangle$, which is sent to zero in the quotient $\mathbb{R}[x]/\langle x^2 + 1 \rangle$.

$\langle x^2 + 1 \rangle$ is the ideal generated by $x^2 + 1$ in the ring $\mathbb{R}[x]$. Since it is singly generated, it is a principal ideal. Since the degree of $x^2 + 1$ is $2$, all polynomials of degree $2$ or greater in $\mathbb{R}[x]$ are reduced to linear or constant representatives of equivalence classes in $\mathbb{R}[x]/\langle x^2 + 1 \rangle$.

The second display in the yellow box shows that $\alpha^2 + 1 \cong 0 \pmod{\langle x^2 + 1 \rangle}$ in $\mathbb{R}[x]/\langle x^2 + 1 \rangle$.

Answer 3

I would recommend you to re-read your definitions because your question is about what an ideal presentation and a quotient ring are. Anyway, since you're asking, I will answer you in a general fashion.

First of all, $\langle f(x)\rangle\subseteq k[x]$ is the set of all polynomials of the form $g(x)f(x)$ with $g(x)\in k[x]$, equipped with the usual operations of the polynomial ring (you can think it as the multiples of the polinomial $f(x)$).

Second, the quotient ring $k[x]/\langle f(x)\rangle$ is the quotient induced by the equivalence relation "$g(x)\sim h(x)$ if and only if $g(x)-h(x)\in \langle f(x)\rangle$". Clearly, the equivalence class of $f(x)$ equals the equivalence class of $0$, since $f(x)-0=f(x)=1\cdot f(x)\in \langle f(x)\rangle $.