# $(R_1\oplus R_2) [x]/(p(x)) = R_1[x]/(p(x))\oplus R_2[x]/(p(x))$?

For convenience, I shall use '$=$' to denote isomorphisms.

Suppose we have a commutative ring $R = R_1\oplus R_2$, and $(p(x))$ is the ideal generated by $p(x)\in R[x]$. Can we deduce that $R[x]/(p(x))=R_1[x]/(p(x))\oplus R_2[x]/(p(x))$ ?

Here, the coefficients of $p(x)$ can be projected into $R_1$ or $R_2$ when it is necessary.

## 2 Answers

Formally speaking, you can't even write $R_i[x]/(p(x))$ unless $p(x)\subseteq R_i[x]$.

But your idea of projecting coefficients does help, though. Splitting on the coordinates, you get $p(x)=p_1(x)+p_2(x)$, and clearly $(p_i(x))\subseteq R_i[x]$.

Then the rings $R_i/(p_i(x))$ make sense, and you can confirm that $p_1(x)R_1$ and $p_2(x)R_2$ are $R$-ideals under the coordinatewise action of $R$, and that

$R[x]/(p(x))\cong R_1[x]/(p_1(x))\oplus R_2[x]/(p_2(x))$

First we have $R[x]\cong R_1[x]\oplus R_2[x]$, the isomorphism being given by $$(a_0,b_0)+(a_1,b_1)x+\cdots+(a_n,b_n)x^n\mapsto (a_0+a_1x+\cdots+a_nx^n,b_0+b_1x+\cdots+b_nx^n).$$ Now if $p(x)=(a_0,b_0)+(a_1,b_1)x+\cdots+(a_n,b_n)x^n\in R[x]$, then the image of the ideal $(p(x))$ is nothing but the direct sum of ideals $(p_1(x))\oplus(p_2(x))$, where $p_1(x)=a_0+a_1x+\cdots+a_nx^n$, $p_2(x)=b_0+b_1x+\cdots+b_nx^n$. So we get $$R[x]/(p(x))\cong R_1[x]/(p_1(x))\oplus R_2[x]/(p_2(x)).$$

Edit. Since I know you are interested in $\mathbb{Z}_6[x]/(x^3-1)$ let's start from $\mathbb{Z}_6\cong\mathbb{Z}_2\oplus\mathbb{Z}_3$. (This isomorphism is given by $a\pmod 6\mapsto(a\pmod 2,a\pmod 3)$.) The image of the polynomial $x^3-1\in\Bbb Z_6[x]$ in $(\mathbb{Z}_2\oplus\mathbb{Z}_3)[x]$ is $(1,1)x^3+(-1,-1)$ which corresponds to $(x^3-1,x^3-1)$ in $\mathbb{Z}_2[x]\oplus\mathbb{Z}_3[x]$, and this is why we have $$\mathbb{Z}_6[x]/(x^3-1)\cong\mathbb{Z}_2[x]/(x^3-1)\oplus\mathbb{Z}_3[x]/(x^3-1).$$

• If this is the case, then there are 16 ideals containing $(6, x^3-1)$ in $\mathbb{Z}[x]$. – booksee Jan 7 '14 at 21:33