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Let's consider the commutative ring $\mathbb{R}[x]/(x^5+x^3)$. We have that $x^5+x^3=x^3(x^2+1)$. So $\mathbb{R}[x]/(x^3(x^2+1)) \simeq \mathbb{C}[x]/(x^3) $.

How can I write the artinian ring $\mathbb{R}[x]/(x^5+x^3)$ as direct product of its localizations?

ArthurStuart
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    $x^3$ and $x^2+1$ are relatively prime. Use the Chinese remainder theorem. So, $\mathbb{R}[x]/(x^3(x^2+1))\simeq\mathbb{R}[x]/(x^3)\otimes\mathbb{R}[x]/(x^2+1)\simeq\mathbb{R}[x]/(x^3)\otimes\mathbb{C}$. – OR. Dec 21 '13 at 19:46
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    @ArthurStuart: Your isomorphism is not correct. @ ABC: You mean direct products. – Martin Brandenburg Dec 21 '13 at 20:06
  • The Chinese remainder theorem gives products, yes. – OR. Dec 21 '13 at 22:10

1 Answers1

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As you already can see from the comments your ring is isomorphic to a direct product via CRT.

Since you asked for a decomposition of the ring as a direct product of localizations I'd like to add few things:

$R=\mathbb R[x]/(x^5+x^3)$ has exactly two prime ideals, $(x)$ and $(x^2+1)$, and therefore two localizations.

We have $R_{(x)}\simeq \mathbb R[x]_{(x)}/(x^5+x^3)\mathbb R[x]_{(x)}$ and since $(x^5+x^3)\mathbb R[x]_{(x)}=x^3\mathbb R[x]_{(x)}$ we get $R_{(x)}\simeq\mathbb R[x]_{(x)}/x^3\mathbb R[x]_{(x)}\simeq (\mathbb R[x]/x^3\mathbb R[x])_{(x)}$. But $\mathbb R[x]/x^3\mathbb R[x]$ is in fact a local ring whose maximal ideal is $(x)$, so $(\mathbb R[x]/x^3\mathbb R[x])_{(x)}=\mathbb R[x]/(x^3)$.

Analogously we get $R_{(x^2+1)}\simeq \mathbb R[x]/(x^2+1)$.

Now we can write $R\simeq R_{(x)}\times R_{(x^2+1)}$.