# Not understanding the isomorphism R x R = C

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?

## Answers and Replies

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.

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
Staff Emeritus