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I've unintentionally "proved" the following: $$\mathbb{C} \cong \mathbb{C} \times \mathbb{C}$$ Can you help me tracing the error I made resulting to this non-proof? Here it is.

First of all, I recall an algebraic theorem about the complex plane: $\mathbb{C} \cong \mathbb{R}[X]/(X^2+1)$. The rest of the non-proof is about the apparent isomorphism $\mathbb{C} \times \mathbb{C} \cong \mathbb{R}[X]/(X^2+1)$.

The ring $R[X]$ is a commutative ring containing a unit, so we can write the ideal $(X^2+1)$ as a product of ideals $(X-i)(X+i)$. These ideals are indivisable: $(X+i)(X-i) \ni \frac{1}{2}(X+i)-\frac{1}{2}(X-i)=i$, so every polynomial $q = i^3q \cdot i$ is contained in the sum of ideals. Now we can use the generalisation of the Chinese remainder theorem: For a commutative ring R with unity, and indivisible ideals $I,J$ applies $R/IJ \cong R/I \times R/J$.

Now we have $\mathbb{R}[X]/(X^2+1) \cong \mathbb{R}[X]/(X+i) \times \mathbb{R}[X]/(X-i)$. I recall that for every homomorphism of rings $f: R \rightarrow S$ the isomorphism $R/\ker(f) \cong f(R)$ holds. Consider the mapping $s_r \ : \ R[X] \rightarrow S \ : \ \sum a_i X^i \mapsto \sum a_i r^i$. This substitution map is an homomorphism. A special case is the homomorphism $s_i \ : \ \mathbb{R}[X] \rightarrow \mathbb{C} \ : \ \sum a_i X^i \mapsto \sum a_i r^i$ that substitutes $i$ in $X$. Its kernel is exactly the ideal $(X-i)$. The image clairy contains $\{ a + bi: a,b \in \mathbb{R} \} = \mathbb{C}$, and the image is contained in \matbb{C} as well. So we obtain $\mathbb{C} \cong \mathbb{R}[X]/(X+i)$. The same trick with the substitution s_{-i} shows that $\mathbb{C} \cong \mathbb{R}[X]/(X-i)$. This results into $\mathbb{C} \cong \mathbb{C} \times \mathbb{C}$. Here ends the "proof."

I feel a little bad about using $s_i$, because $i \notin \mathbb{R}$, but I remember that a similar mapping had to be used to proof that $\mathbb{C} \cong \mathbb{R}[X]/(X^2+1)$.

I'd be thankful if you help me to sift this through and find the error(s).

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2  
Why the downvote? –  user1729 Aug 1 '13 at 20:44
10  
By the way, you should always specify what type of isomorphisms you mean. For example, how people would have answered why this couldn't be true depends on the isomorphism probably. If you mean isomorphism of $\mathbb{C}$-algebras/spaces people would have told you to count dimensions, if you had said of rings people may have told you to look for zero divisors, and if you had said as groups people would have told you that the isomorphism you wrote is correct. –  Alex Youcis Aug 1 '13 at 20:49
    
@AlexYoucis +1 for detail and completeness. –  Pedro Tamaroff Aug 1 '13 at 21:05

3 Answers 3

up vote 13 down vote accepted

It's not too tough to spot! $i\notin\mathbb R$.

What your proof does in fact show is that by Chinese Remainder, $\mathbb C[X]/(X^2+1)\cong \mathbb C\times\mathbb C$.

The kernel of your homomorphism is $(X^2+1)$, not $(X-i)$, since $X-i\notin\mathbb R[X]$.

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1  
+1, More specifically, $x^2+1$ is irreducible in $\mathbb{R}[x]$. –  Alex Youcis Aug 1 '13 at 20:44
    
Indeed. Which is why the quotient is a field in the first place. :) –  Ted Shifrin Aug 1 '13 at 20:48

What does $\mathbb{R}[X]/(X+\operatorname{i})$ mean? The polynomial $X+\operatorname{i}$ does not belong to $\mathbb{R}[X]$, and so $\mathbb{R}[X]/(X+\operatorname{i})$ does not make sense. This is because $\operatorname{i}$ does not belong to $\mathbb{R}$.

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$ \Bbb R[x]/(x+i)$ is not defined as $x+i$ is not even in $ \Bbb R[x]$.

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