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Let $\mathbb{C}=\{a+bi | a,b \in \mathbb{R}\}$, denote the set of complex numbers, where $i$ has the property that it's square is $-1$.

We define the two binary operation, $+, *$ on this set $\mathbb{C}$, namely addition and multiplication of complex numbers as follows $$(a+bi)+(c+di)=(a+c)+(b+d)i$$ and $$(a+bi)*(c+di)=(ac-bd)+(ad+bc)i$$

We observe that $(\mathbb{C},+,*)$ forms a mathematical object called field.

Note that $*$ has "usual" properties like distributivity, associativity and commutativity.

I have the following Question, Are there many ways of defining binary operation $*$ (or call it multiplication of complex number ), such that these above mentioned properties are satisfied.

Tensor_Product
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    Not that makes a field (there is, up to change of basis, only one way to make a field with the standard addition), but if you just want a ring (distribution and association), then there are several ways to do it. For instance, you could define $(a+bi)(c+di)=ac+bdi$. That's otherwise known as $\Bbb R^2$. You could also define the trivial multiplication: Every multiplication results in $0$. This is not allowed of your ring theory demands, for instance, that rings are unital with $1\neq0$, but it is associative and distributive. – Arthur Feb 08 '17 at 17:39
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    The most obvious way is $(a+bi)\otimes(c+di)=ac+bdi$. This is also associative, commutative, and distributive, but it has the disadvantage that there are zero divisors - $(a+0i)(0+di)=0$. – Thomas Andrews Feb 08 '17 at 17:40
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    Take any field $F$ with $2^{\aleph_0}$ elements, and choose a bijective map $F \to \Bbb C$. Then it induces a field structure on $\Bbb C$. But if you take $F = \Bbb R$, then this field structure gives a non-algebraically closed field, which is not isomorphic to the usual field structure on $\Bbb C$. – Watson Feb 08 '17 at 17:42
  • @Watson That would also yield a different addition, which as I read it, is not what is wanted. – Arthur Feb 08 '17 at 17:43
  • @Arthur There are other ways to get a field, but they are isomorphic. – Thomas Andrews Feb 08 '17 at 17:43
  • @ThomasAndrews I did say that there is only one way "up to change of basis". – Arthur Feb 08 '17 at 17:48
  • @Arthur, If I drop the condition of "standard addition", then what happens, Do we get a different field? any example. – Tensor_Product Feb 08 '17 at 17:49
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    (Related for $\Bbb R$: https://math.stackexchange.com/questions/105446, https://math.stackexchange.com/questions/985876, https://math.stackexchange.com/questions/1626926). – Watson Feb 08 '17 at 17:49
  • @Arthur: You're assuming that the result has to be an $\mathbb{R}$-algebra in the obvious way. –  Feb 08 '17 at 17:51
  • @Tensor_Product If you want to build this from the bottom up including addition, then the only thing you're stuck with is the cardinality. In that case, you can get any field of cardinality $|\Bbb R|$ you want, from $\Bbb R$ to the field of rational functions and beyond. – Arthur Feb 08 '17 at 17:51
  • @Hurkyl If there is another field using the addition structure on $\Bbb C$ without that requirement, I would love to hear about it. – Arthur Feb 08 '17 at 17:52
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    Also related: http://math.stackexchange.com/questions/526540, http://math.stackexchange.com/questions/1415915, http://math.stackexchange.com/questions/394247 – Watson Feb 08 '17 at 21:59
  • To the most upvoted comments, (a+bi)(c+di)=ac+bdi wouldn't make i squared equal to minus 1. So they are wrong. – FCardelle Jun 07 '21 at 19:49

2 Answers2

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The underlying additive group of the field $\mathbb{C}$ is just a $\mathbb{Q}$-vector space of dimension $\mathfrak{c}$.

By transport of structure, we can define a multiplication on $\mathbb{C}$ to make it isomorphic to any other such structure whose additive group is a $\mathbb{Q}$-vector space of dimension $\mathfrak{c}$.

One way to identify such is that they are precisely the $\mathbb{Q}$-algebras of cardinality $\mathfrak{c}$.

Examples of fields that we can make the result isomorphic to include:

  • The real numbers $\mathbb{R}$
  • The $p$-adic numbers $\mathbb{Q}_p$
  • The field of rational Laurent series $\mathbb{Q}((x))$
  • The function field $F(x)$ where $F$ is any field appearing in this list

And I've only stuck to relatively simple examples. Lots more possibilities open up if we only require the structure to be a commutive ring, or even merely a commutative $\mathbb{Q}$-algebra.

You might want to consider requiring the end result be a $\mathbb{R}$-algebra in the obvious way: that if $r$ is a real number, then $(r + 0i) * (a+bi) = (ra) + (rb)i$.

In this case, there are only three possibilities, up to isomorphism of $\mathbb{R}$-algebras.

  • The complex numbers (e.g. $i^2 = -1$)
  • The dual numbers (e.g. $i^2 = 0$)
  • The hyperbolic numbers (e.g. $i^2 = 1$)

The hyperbolic numbers are isomorphic to $\mathbb{R} \times \mathbb{R}$.

Given a multiplication operation, which case you are in can be determined by looking at the minimal polynomial of $i$. Its discriminant will either be negative, zero, or positive respectively.

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Let's not use $i$, because it is a distraction.

Given $x,y$ with $y\neq 0$, you can define:

$$(a+b\alpha)\otimes(c+d\alpha)= (ac-bd(x^2+y^2))+(ad+bc+2bdx)\alpha$$

This operation, when $x=0,y=1$, agrees with your operation.

This operation is basically gotten by thinking of $\alpha=x+yi$. Then $\alpha^2=2x\alpha -(x^2+y^2)$.

It is commutative, associative and distributive even if $y=0$. It just has zero divisors in that case. For example, if $x=1,y=0$, then $(1-\alpha)^2=0$.

When $y\neq 0$, it defines a field that is isomorphic with the complex numbers, but appears quite different.

More generally, given any real $u,v$ you can define:

$$(a+b\alpha)\otimes(c+d\alpha)=(ac+bdu)+(ad+bc+vbd)\alpha$$

These are all the cases of commutative, associative and distribute multiplications where we have that $(a+0\alpha)\otimes(c+d\alpha)=ac+ad\alpha$, so that we can think of our new ring as an extension of the real numbers.

Thomas Andrews
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