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The question is in the title. By the first isomorphism theorem, I know that if I can find a surjective ring homomorphism $\varphi : \mathbb{R}[x] \rightarrow S$, where $S$ is some standard ring, and ker$(\varphi) = (x^3 +x)$, then I am set.

So I chose $\varphi : \mathbb{R}[x]\rightarrow\mathbb{R}$, where $\varphi (p(x)) = p(0)$ for some $p(x)\in\mathbb{R}[x]$. However, I don't think this map satisfies my kernel requirement.

My question is how can I adjust my mapping to account for ker$(\varphi) = (x^3 +x)$?

Thank you! ~Dom

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3 Answers

up vote 6 down vote accepted

Hint: you should use the Chinese remainder theorem for $\def\Kx{K[x]}\Kx$ here ($K$ a field): just like for $\Bbb Z$ one has $\def\Z{\Bbb Z}\Z/n\Z\simeq (\Z/a\Z)\times(\Z/b\Z)$ when $n=ab$ is a decomposition of $n$ into a product of relatively prime numbers $a,b$, one has $\Kx/(P)\simeq(\Kx/(A))\times(\Kx/(B))$ when $P=AB$ is a decomposition of $P$ into a product of relatively prime polynomials $A,B$.

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$\rm h : \Bbb R[x]\to \Bbb R \times \Bbb C\:$ via $\rm\: f(x)\mapsto (f(0),f({\it i}\,))\,$ is onto: $\rm\:a\!+\!(a\!-\!b)x^2\mapsto (a,b).\:$ So $\rm\,Im\ h = \Bbb R \times \Bbb C.\,$ Next, $\rm\:f\in ker\ h\!\iff\!x,\,x^2\!+1\mid f\!\iff\!lcm(x,x^2\!+1) = x^3\!+x\mid f,\:$ thus $\rm\:ker\ h = (x^3+x).\:$ Applying the First Isomorphism Theorem $\rm\:R[x]/(ker\ h) \cong\, Im\ h,\:$ i.e. $\rm\:R[x]/(x^3\!+x)\cong \Bbb R \times \Bbb C$.

Remark $\ $ Weierstrass (1884) and Dedekind (1885) showed that every finite dimensional commutative extension ring of $\mathbb R$ without nilpotents $\rm (x^n = 0 \ \Rightarrow\ x = 0)$ is isomorphic as a ring to a direct sum of copies of $\rm\:\mathbb R\:$ and $\rm\:\mathbb C\:.\:$ Wedderburn and Artin proved a generalization that every finite-dimensional associative algebra without nilpotent elements over a field $\rm\:F\:$ is a finite direct sum of fields. See this answer for much more on this and related matters.

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Since $\gcd(x,x^2+1)=1$, then by the Chinese remainder theorem

we have $$\mathbb{R}[x]/(x^3+x)\simeq\mathbb{R}[x]/x\times \mathbb{R}[x]/(x^2+1).$$ Moreover, we have $$\mathbb{R}[x]/x\simeq \mathbb{R},$$ and $$\mathbb{R}[x]/(x^2+1)\simeq\mathbb{C}\simeq\mathbb{R}^2,$$ so, we finally find $$\mathbb{R}[x]/(x^3+x)\simeq\mathbb{R}\times \mathbb{C}\simeq\mathbb{R}^3.$$

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Are you sure that $\mathbb C\simeq\mathbb R^2$? –  user26857 Mar 6 '13 at 8:54
    
@YACP If we define the multiplication in $\mathbb{R}^2$ by $$(a,b)\times(a',b')=(aa'-bb',ab'+a'b),$$ then we can find a ring isomorphism. Is not it? –  Sami Ben Romdhane Mar 6 '13 at 9:14
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Nice try! Nobody can think of $\mathbb R^2$ with this multiplication, and even less (if possible to say so) to $\mathbb R^3$ as being $\mathbb R\times\mathbb R^2$ where $\mathbb R$ is taken with the usual multiplication and $\mathbb R^2$ with that one which gives an isomorphism with $\mathbb C$. –  user26857 Mar 6 '13 at 9:20
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