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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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All the laws with fancy names are valid for complex numbers. But that is not what the fundamental Theorem of Algebra says. –  André Nicolas Mar 24 '12 at 19:14
    
Do you go to UCLA? If so, are you talking about $115B$? –  Daniel Montealegre Mar 24 '12 at 21:00
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It's incredible that a math teacher would claim the Fundamental Theorem of Algebra is the same thing as the existence of the four basic operations on complex numbers extending similar operations on real numbers. Does he think the Fundamental Theorem of Calculus asserts that a continuous function on a closed and bounded interval has a maximum value? That would make about as much (non)sense. –  KCd Mar 24 '12 at 22:21
    
@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. –  Neo Mar 24 '12 at 23:21
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2 Answers

up vote 4 down vote accepted

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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Yes that is precisely what I thought. However, am I right in stating that in the above statements, statement two directly leads to statement one? –  Neo Mar 24 '12 at 19:12
    
Not quite; for example the quaternions contain a copy of the real numbers, they have an addition, subtraction, multiplication, and division which extend the corresponding operations on the real numbers, but the commutativity of multiplication is not true for general quaternions. So just having operations which extend the operations on the reals doesn't imply that they will continue to have the same properties. –  Zev Chonoles Mar 24 '12 at 19:19
    
I'm sorry. I meant does statement one lead to statement two? My apologies. –  Neo Mar 24 '12 at 19:25
    
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. –  André Nicolas Mar 24 '12 at 19:43
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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. –  Neo Mar 24 '12 at 23:32
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