Not understanding the isomorphism R x R = C

[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?

 

[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.

 

[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!

 

[Fredrik]

I'd always imagined isomorphisms as carrying over all structure, including invertibility.

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 $(a,b)(c,d)=(ac-bd,ad+bc)$.