2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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 $$\mathbb{Z}_6[x]/(x^3-1)\cong\mathbb{Z}_2[x]/(x^3-1)\oplus\mathbb{Z}_3[x]/(x^3-1).$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.