Not understanding the isomorphism R x R = C

  • #1
53
6

Main Question or Discussion Point

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

  • #2
177
61
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
53
6
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
Fredrik
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Science Advisor
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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)##.
 

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