Is the complex space $\mathbb C\times \mathbb C$ a field?

Question

Is the complex space $\mathbb C\times \mathbb C$ a field?
If not, which axiom that defines field doesn't hold?

Answer 1

Presume you are performing addition and multiplication component-wise, i.e.

$$(x_1, y_1)+(x_2, y_2)=(x_1+y_1, x_2+y_2)$$ $$(x_1, y_1)(x_2, y_2)=(x_1y_1, x_2y_2)$$

In that case, the role of multiplicative unity ("one") is played by $(1,1)$:

$$(1,1)(x,y)=(1x, 1y)=(x,y)$$

and the role of additive unity ("zero") is played by $(0,0)$:

$$(0,0)+(x,y)=(0+x, 0+y)=(x,y)$$

However, look at the element $(1,0)$. It is not zero, but it hasn't got a multiplicative inverse, because there is no pair $(x,y)$ such that $(1,0)(x,y)=(1,1)$ - simply because $0y=0\ne 1$ for every $y$.

So, $\mathbb C\times\mathbb C$ is not a field because there are non-zero elements which don't have a multiplicative inverse.

Answer 2

Just to add to @StinkingBishop's answer, the reason it starts with "Presume you are adding..." is this:

The cardinality of $\Bbb C \times \Bbb C$ is the same as that of $\Bbb R$, i.e., there exists some bijection $f$ from the first to the second, with inverse I'll call $g$ to save typing. If you define $$ (a,b) \oplus (c, d) = g\left( f(a, b) + f(c, d) \right) $$ and $$ (a,b) \odot (c, d) = g\left( f(a, b) \cdot f(c, d) \right) $$

then $\Bbb C \times \Bbb C$ actually is a field with the addition and multiplication operations $\oplus$ and $\odot$.

The broader implication of this contrived example is that it generally doesn't make sense to ask "Is such-and-such a set actually a field?" without specifying the operations you're considering.

There are special cases, however, where an intermediate kind of question makes sense. You can ask "Is there a multiplication operation $\otimes: S^2 \times S^2 \to S^2$ on the 2-sphere that makes the 2-sphere into a group, and in which $\otimes$ is a differentiable function?" In this case, you haven't specified what the operation is, but you have put some restriction on it. (Answer: it turns out that there's no such "multiplication", but it's not all that easy to prove.)

Answer 3

John Hughes, I see what you mean. You are basically re-writing the elements of the set $\mathbb R$ using the elements of the set $\mathbb C\times \mathbb C$ (which is possible as they have the same cardinality). Then you multiply and add the new elements just as if they still were in $\mathbb R$. This clearly makes $\mathbb C\times \mathbb C$ a field with respect to new fancy operations.

But I am confident Karen was meaning exactly what Stinking Bishop assumed. If I were an instructor who plans to include this in homework, I would necessarily mention that point.