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We had a quiz recently in a linear algebra course, and one of the true/false question states that

The Fundamental Theorem of Algebra asserts that addition, subtraction, multiplication and division for real numbers can be carried over to complex numbers as long as division by zero is avoided.

According to our teacher, the above statement is true. When asked him of the reasoning behind it, he said something about the FTA asserts that the associative, commutative and distributive laws are valid for complex numbers, but I couldn't see this. Can someone explain whether the above statement is true and why? Thanks.

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    @DanielMontealegre Haha no I'm a high school student. I had a long discussion with him and he said he got the question from a book somewhere. I guess I'll read through the book once he finds it and see what the book says.2012-03-24

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The statement is false.

The Fundamental Theorem of Algebra asserts that any non-constant polynomial with complex coefficients has a root in the complex numbers. This does not state anything about the relationship between the complex numbers and the real numbers; and any proof of the FTA will certainly use the associativity and commutativity of addition and multiplication in the complex numbers, as well as multiplication's distributivity over addition, so the FTA can't imply those properties.

The statements

  • the associative, commutative and distributive laws are valid for complex numbers
  • addition, subtraction, multiplication and division for real numbers can be carried over to complex numbers as long as division by zero is avoided

might be summarized by the statement "the complex numbers form a ring, which is a division algebra over the real numbers".

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    Statement $1$ does not lead to statement $2$. By the very imprecise phrase "can be carried over" what is presumably meant is that the field of real numbers is a subfield of the field of complex numbers, meaning that for example that when we take two complex numbers $x$ and $y$ which happen to be real, the rule for multiplication of $x$ and $y$ as complex numbers gives the same answer as the ordinary product of $x$ and $y$ in the reals.2012-03-24
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What your instructor probably had in mind is the following.

Let $z_1=a_1+ib_1$ and $z_2=a_2+ib_2$ be two complex numbers.

Then, the polynomials $P(X)= [\frac{1}{2}X+a_1]+[\frac{1}{2}X+a_2]+i(b_1+b_2)$ has by FTA a complex root, which we can define as $z_1+z_2$.

Similar things can be done for $z_1z_2$ and $\frac{1}{z_1}$.

Anyhow, given a polynomial $P$, even if somehow you can define the polynomial algebraically only by using real operations and $i$, you cannot use the FTA without defining first addition and multiplication. The FTA is not about the polynomial as an algebraic expression, it is about the Polynomial as a function...

To Quote zev, FTA asserts that any non-constant polynomial with complex coefficients has a root in the complex number. But what does a root of a polynomial mean? How do you calculate/evaluate $P(z)$ if you don't know how to calculate the powers of $z$, and multiply $z$ to the coefficients of the polynomial? And how do you add the monomials of the polynomial together?

The only way in which you can get around this issue is by trying to define $C$ as the algebraic closure of $R$, and then the algebraic process of "algebraic closure" and the uniqueness of it shows that any algebraic extension of $R$ in which FTA holds has to be $C$... But then, if you use this definition of $C$, how do you prove that $C= \{ a+bi |a,b \in R\}$??

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    Right, and that is exactly what I told my teacher. Logically the complex numbers and their algebraic operations would be well established before the FTA came about. You cannot say that simply because a polynomial may have complex roots, these complex numbers behave the same way as real numbers. Thank you for clearing that up.2012-03-24