In my complex analysis lectures, the following definition (translated) is given for $\mathbb{C}$:

A complex plane $\mathbb{C}$ is a number plane $\mathbb{R}^2$ with the operations of modulus, addition and multiplication as defined before.

EDIT: the modulus operation is defined on $\mathbb{R}^2$: $$(x,y) \in \mathbb{R}^2 \rightarrow |x+iy| = \sqrt{x^2+y^2}. $$

I do not doubt that a properly defined field $\mathbb{C}$ would possess the modulus operation. However, is the modulus operation necessary to properly define $\mathbb{C}$? What if I define $\mathbb{C}$ as

a number plane $\mathbb{R}^2$ with the operations of addition and multiplication as defined before.

Can another field $\mathbb{D}$ possess the same operations of addition and multiplication as $\mathbb{C}$, but a different modulus operation and a different field structure?

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    $\begingroup$ What does "modulus operation" mean to you, precisely? $\endgroup$ – Stahl Jun 17 '17 at 19:59
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    $\begingroup$ What does "field structure" mean, if not the operations of addition and multiplication? The modulus makes $\mathbb{C}$ into a normed space and a metric space, so it provides topological information. $\endgroup$ – André 3000 Jun 17 '17 at 20:04
  • $\begingroup$ With the addition and multiplication you get the algebraic structure. Maybe they want to make clear the topology by saying that it comes from the norm $\endgroup$ – OR. Jun 17 '17 at 20:04
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    $\begingroup$ The data of a field is $(k, +, \cdot)$, where $k$ is a set and $+,\cdot : k\times k\to k$ are two binary operations on $k$ subject to some axioms: there is no modulus involved. So if you have another field which "has the same" multiplication and addition as $\Bbb C$, it has to be [isomorphic to] $\Bbb C$, because that's all the data that is encoded in the field structure of $\Bbb C$. $\endgroup$ – Stahl Jun 17 '17 at 20:07
  • $\begingroup$ @Stahl added the modulus definition to the post. $\endgroup$ – svavil Jun 17 '17 at 20:29

Definition: A field is a set $k$ with two binary operations $+,\cdot : k\times k\to k$ satisfying the following axioms:

  1. For all $a,b,c\in k$, $(a + b) + c = a + (b + c)$.
  2. For all $a,b\in k$, $a + b = b + a$.
  3. There exists $0\in k$ such that $a + 0 = a$ for all $a\in k$.
  4. For every $a\in k$, there exists $a'\in k$ such that $a + a' = 0$.
  5. For all $a,b,c\in k$, $(a\cdot b)\cdot c = a\cdot (b\cdot c)$.
  6. For all $a,b\in k$, $ab = ba$.
  7. There exists $1\in k\setminus\{0\}$ such that $1\cdot a = a$ for all $a\in k$.
  8. For all $a,b,c\in k$, $a\cdot(b + c) = a\cdot b + a\cdot c$.
  9. For all nonzero $a\in k$, there exists $\tilde{a}\in k$ such that $a\cdot\tilde{a} = 1$.

Note that there is no "modulus" involved in the above definition. This is all a very long-winded way to say that $\Bbb C$ as a field is essentially only determined by its addition and multiplication, so that the answer to "can a field $\Bbb D$ possess the same operations of addition and multiplication as $\Bbb C$, but a different field structure?" is no. There are many fields which are algebrically "the same" as $\Bbb C$ (that is, they are isomorphic as fields). Here are a few:

  • $(\Bbb C,+,\cdot)$, where $\Bbb C = \{a + bi\mid a,b\in\Bbb R\}$ with the "obvious" addition and multiplication subject to the rule that $i^2 = -1$,
  • $(\Bbb R^2, + , \cdot)$, where $(a,b) + (c,d) = (a + c, b + d)$ and $(a,b)\cdot (c,d) = (ac - bd, ad + bc)$,
  • the algebraic closure of the field $\Bbb R$ of real numbers,
  • $\Bbb R[x]/(x^2 + 1)$,
  • the unique (up to isomorphism) algebraically closed field of characteristic $0$ with cardinality $2^{\aleph_0}.$

However, it is possible to put different notions of modulus on the same field. As mentioned in the comments, the modulus is providing topological/geometric information (it is essentially giving a way to measure distance in your field), not algebraic information. Let us define a modulus on a field in general:

Definition: Let $(k,+,\cdot)$ be a field. Then a modulus on $k$ is a function $v : k\to\Bbb R$ such that

  1. $v(a)\geq 0$ for all $a\in k$,
  2. $v(a) = 0$ if and only if $a = 0$,
  3. $v(ab) = v(a)v(b)$ for all $a,b\in k$,
  4. $v(a + b)\leq v(a) + v(b)$ for all $a,b\in k$.

The distance between two points $a,b\in k$ can then be defined as $v(a - b)$, and you can check that this satisfies the usual properties one would expect of a notion of distance. We can now talk about valued fields: these are fields with a choice of modulus on that field; i.e., a valued field is the data of $(k,+,\cdot,v)$, where $(k,+,\cdot)$ is a field, and $v$ is a modulus on $k$.

It is possible to have two valued fields whose underlying algebraic field structure is the same, but whose moduli are not equivalent. For example, take the first valued field to be $\Bbb C$ with the usual modulus, as you've defined, and take the second to be $\Bbb C$, but now with the following modulus: $v(a + bi) = 1$ for all nonzero $a + bi\in\Bbb C$, and $v(0) = 0$. This is kind of a silly modulus, but it should be intuitively obvious that this is different than the normal modulus on $\Bbb C$. There are other more complicated examples as well. You can also look here for a more thorough discussion, or just google search "absolute value on a field."

When one speaks of the complex numbers, one usually means $\Bbb C$ as a valued field, not just as a field, because the geometric/topological structure determined by the usual modulus on $\Bbb C$ is very important. Complex geometry and complex analysis are very rich subjects with many beautiful results that use the geometry coming from the usual modulus on $\Bbb C$ in a crucial way. These results are generally not transferrable to valued fields whose underlying field is $\Bbb C$ but have a different modulus.

  • $\begingroup$ Great answer! __ $\endgroup$ – Vincent Jun 17 '17 at 21:31
  • $\begingroup$ (Apologies for the stupid __ in my previous comment. Apparently comments need to be 15 characters long) $\endgroup$ – Vincent Jun 17 '17 at 21:32

I am guessing that complex analysis lecturer wants $\mathbb C$ to be more than merely an algebraically closed field with power $2^{\aleph_0}$. They want also the topological structure given by the modulus. Note that for the real numbers, we have $x>y$ definable in terms of multiplication and addition, so we can define the topological structure from the field structure. Not so for the complex numbers.


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