# Not understanding the isomorphism R x R = C

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1. Feb 1, 2015

### metapuff

Now ℝxℝ≅ℂ, seen by the map that sends (a,b) to a + bi. ℂ is a field, so the product of any two non-zero elements is non-zero. However, this doesn't seem to hold in ℝxℝ, since (1,0) * (0,1) = (0,0) even though (1,0) and (0,1) are non-zero. What am I missing?

Also, the zero ideal is maximal in ℂ, since ℂ is a field. But in ℝxℝ, we have two maximal ideals: ℝx{0} and {0}xℝ. Surely the isomorphism between ℝxℝ and ℂ ought to preserve the ideal structure?

2. Feb 1, 2015

### Hawkeye18

Isomorphism between $\mathbb R\times\mathbb R$ and $\mathbb C$ is an isomorphism between real vector spaces, i.e. it agrees only with addition and multiplication by real numbers. But is does not agree with the multiplication: the "natural" coordinatewise multiplication in $\mathbb R\times\mathbb R$ does not make it a filed, it only gives you a ring. The multiplication in $\mathbb C$ is quite different, and it makes $\mathbb C$ a field.

3. Feb 1, 2015

### metapuff

Ah yeah, you're right. There's no square root of -1 in ℝxℝ anyway, so it's no surprise that multiplication doesn't carry over. I'd always imagined isomorphisms as carrying over all structure, including invertibility. Thanks for clearing this up!

4. Feb 12, 2015

### Fredrik

Staff Emeritus
Isomorphisms do, but not all bijective maps are isomorphisms. The specific map you mentioned is linear. That makes it a vector space isomorphism, if we view $\mathbb C$ as a vector space over $\mathbb R$, but it's not a ring isomorphism (and therefore not a field isomorphism) unless you have chosen the product operation on $\mathbb R\times\mathbb R$ to be something like the one defined by $(a,b)(c,d)=(ac-bd,ad+bc)$.