I know that it is not a PID since $((x-1)(x+1) + I)(x+3)(x-5) + I) = 0$ where $I$ is the ideal we are quotienting by, and this means it is not even an Integral Domain since it has zero divisors different than zero. However, I was wondering if one could use the Chinese Remainder Theorem to show that, since all of $(x-1),\ (x+1),\ (x+3),\ (x-5)$ are pairwise comaximal (subtract them and obtain a unit in $\mathbb{R}$), then the quotient ring is
$$\mathbb{R}[x]/(x-1)(x+1)(x+3)(x-5) \cong \mathbb{R}[x]/(x-1) \times \mathbb{R}[x]/(x+1) \times \mathbb{R}[x]/(x+3) \times \mathbb{R}[x]/(x-5)$$
But each of these terms satisfy $\mathbb{R}[x]/(x-b) \cong \mathbb{R}[b] \cong \mathbb{R}$, and thus the whole thing is just isomorphic to $\mathbb{R}^4$. However, what even is the ring structure of $\mathbb{R}^4$? Addition can be defined component-wise, but what about multiplication? Does this seem correct or am I missing something?
Best Answer
Addition and multiplication in $\mathbb R^4$ are given pointwise, as in any product of rings.
Let's start with $\mathbb R[x]/(x-a) \cong \mathbb R[a]\cong \mathbb R$ first. By Euclidean division, we can uniquely write any real polynomial as $$p(x) = q(x)(x-a)+r.$$ If we want to sum $p_1(x)$ and $p_2(x)$ in $\mathbb R[x]$, we get $$p_1(x)+p_2(x) = (q_1(x)+q_2(x))(x-a) + (r_1+r_2)$$ and to multiply them, we get $$p_1(x)p_2(x) = (q_1(x)q_2(x)(x-a) + r_1q_2(x)+r_2q_1(x))(x-a)+r_1r_2.$$
From here it's clear that we have $$p_1(x)+p_2(x)\equiv p_1(a)+p_2(a)\pmod{(x-a)}$$ $$p_1(x)p_2(x)\equiv p_1(a)p_2(a)\pmod{(x-a)}$$ and the isomorphism is given by $p(x)+\langle x - a\rangle \mapsto p(a)$.
Now, if we look at $\mathbb R[x]/(x-a)(x-b)\cong \mathbb R[x]/(x-a)\times \mathbb R[x]/(x-b) \cong \mathbb R^2$, for $a\neq b$, we can similarly write $$p(x) = q(x)(x-a)(x-b) + r(x) = q(x)(x-a)(x-b)+\color{blue}{\alpha}\frac{x-b}{a-b}+\color{red}{\beta}\frac{x-a}{b-a}$$ where $\alpha$ and $\beta$ are unique real numbers because $x-a$ and $x-b$ are linearly independent over $\mathbb R$ and therefore form a basis for real polynomials of degree at most $1$. In fact, $\alpha = p(a)$ and $\beta = p(b)$.
Then, we have $$p_1(x)+p_2(x)=(\color{blue}{\alpha_1+\alpha_2})\frac{x-b}{a-b}+(\color{red}{\beta_1+\beta_2})\frac{x-a}{b-a}+(x-a)(x-b)(q_1(x)+q_2(x))$$ $$p_1(x)p_2(x)=\color{blue}{\alpha_1\alpha_2}\left(\frac{x-b}{a-b}\right)^2+\color{red}{\beta_1\beta_2}\left(\frac{x-a}{b-a}\right)^2+(x-a)(x-b)\left(q_1(x)q_2(x)(x-a)(x-b)+(\alpha_2 q_1(x)+\alpha_1q_2(x)))\frac{x-b}{a-b}+ (\beta_2 q_1(x)+\beta_1q_2(x))\frac{x-a}{b-a}-\frac{\alpha_1\beta_2 + \alpha_2\beta_1}{(a-b)^2}\right)$$
Now comes the amusing part: $$x-b = (x-a)+(a-b)\implies\frac{x-b}{a-b} = \frac{x-a}{a-b}+1\implies\frac{x-b}{a-b} \equiv 1 \pmod{(x-a)}$$
and
\begin{align} \frac{x-b}{a-b} = \frac{x-a}{a-b}+1 &\implies \left(\frac{x-b}{a-b}\right)^2 = \frac{(x-a)(x-b)}{(a-b)^2}+\frac{x-b}{a-b}\\&\implies \left(\frac{x-b}{a-b}\right)^2 \equiv \frac{x-b}{a-b} \pmod{(x-a)(x-b)} \end{align} and analogously $$\frac{x-a}{b-a} \equiv 1 \pmod{(x-b)},\quad \left(\frac{x-a}{b-a}\right)^2 \equiv \frac{x-a}{b-a} \pmod{(x-a)(x-b)}.$$
Therefore, we can write $$p_i(x)\equiv\color{blue}{\alpha_i}\frac{x-b}{a-b}+\color{red}{\beta_i}\frac{x-a}{b-a} \pmod{(x-a)(x-b)}$$ $$p_1(x)+p_2(x)\equiv(\color{blue}{\alpha_1+\alpha_2})\frac{x-b}{a-b}+(\color{red}{\beta_1+\beta_2})\frac{x-a}{b-a} \pmod{(x-a)(x-b)}$$ $$p_1(x)p_2(x)\equiv\color{blue}{\alpha_1\alpha_2}\frac{x-b}{a-b}+\color{red}{\beta_1\beta_2}\frac{x-a}{b-a} \pmod{(x-a)(x-b)}$$
This should convince you of the isomorphism $\mathbb R[x]/(x-a)(x-b) \cong \mathbb R^2$, because if you choose basis $\{\frac{x-b}{a-b}+\langle (x-a)(x-b)\rangle, \frac{x-a}{b-a}+\langle (x-a)(x-b)\rangle \}$ for $\mathbb R[x]/(x-a)(x-b)$, you have linear isomorphism by sending it to canonical basis for $\mathbb R^2$ and by the above that linear isomorphism is also multiplicative.
We can also break it in steps via Chinese remainder theorem:
$$p_1(x)+p_2(x) \equiv (p_1(a)+p_2(a))\frac{x-b}{a-b} \equiv \color{blue}{p_1(a)+p_2(a)}\pmod{(x-a)}$$ $$p_1(x)+p_2(x) \equiv (p_1(b)+p_2(b))\frac{x-a}{b-a} \equiv \color{red}{p_1(b)+p_2(b)}\pmod{(x-b)}$$ $$p_1(x)p_2(x) \equiv p_1(a)p_2(a)\left(\frac{x-b}{a-b}\right)^2 \equiv \color{blue}{p_1(a)p_2(a)}\pmod{(x-a)}$$ $$p_1(x)p_2(x) \equiv p_1(b)p_2(b)\left(\frac{x-a}{b-a}\right)^2 \equiv \color{red}{p_1(b)p_2(b)}\pmod{(x-b)}$$ This establishes, by hand, that we have isomorphisms $$p(x)+\langle (x-a)(x-b) \rangle \mapsto (p(x) + \langle x-a \rangle, p(x) + \langle x-b \rangle) \mapsto (\color{blue}{p(a)},\color{red}{p(b)}).$$
Hopefully, this is convincing enough that you agree that this works for $\mathbb R^4$ as well, but if not, you can generalize all of the above by writing
$$p(x) = q(x)\prod_{j=1}^n(x-a_j)+\sum_{i=1}^np(a_i)\prod_{j\neq i}\frac{x-a_j}{a_i-a_j},\ a_i\neq a_j,i\neq j.$$
This last form should remind you of Lagrange interpolation. It tells us that for any distinct points $\{a_1,\ldots,a_n\}$ there is a unique polynomial $p$ of degree (at most) $n-1$ such that $p(a_i) = y_i$, for given values $y_i$. This polynomial gives us the inverse to the above isomorphism $$(y_1,\ldots,y_n)\mapsto p(x) + \langle (x-a_1)\ldots (x-a_n) \rangle \colon \mathbb R^n \to \mathbb R[x]/(x-a_1)\ldots (x-a_n).$$