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My textbook says that fields are closed under multiplication and addition, but isn't multiplication the same as addition? Or is that just for the case of how we learn math in grade school?

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Multiplication of positive integers is repeated addition, but try to interpret $\pi\times e$ as repeated addition. –  Michael Albanese Sep 9 '13 at 22:38
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"isn't multiplication the same as addition" ??? –  Martin Brandenburg Sep 9 '13 at 22:38
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Multiplication is ain't repeat addition, multiplication ain't gonna die. –  Git Gud Sep 9 '13 at 22:39
    
The operations your textbook refers to are just two abstract operations that are traditionally called “addition” and ”multiplication”, because they share many of the properties of addition and multiplication on rational or real numbers, as stated in the axiom for fields. –  egreg Sep 9 '13 at 22:40

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In the postive integers, multiplication can be viewed as repeated addition. In fact, this is almost certainly the way humans first thought of multiplication when it was invented. Carrying this to all integers (including negative integers) is a bit trickier, especially for one negative times another, but one can make sense of it.

But there are simple fields where this thinking doesn't work. For example, take the ring of all polynomials over $\mathbb{Z}_2$. (In other words, the ring of all polynomials with coefficients 0 and 1.) Now mod out by $x^2+x+1$, which is irreducible over $\mathbb{Z}_2$. The resulting field has four elements:

$$ 0, 1, x, 1+x $$

Addition is normal polynomial addition where the coefficients are mod 2, so any element added to itself is 0. (You can think of this as XOR if you're a CS person.) But multiplication is multiplication mod $x^2+x+1$, which cannot be thought of as repeated addition since it's not even clear how to interpret $x$ times $x+1$ as repeated addition. (You don't need a field to realize this..., but since that's the context you asked about).

In case you're interested, $x \cdot (x+1) = 1$ in this field. But if you take (say) $x$ and repeatedly add it to itself (in an attempt to multiply), you will get $x, 0, x, 0, x, \cdots$ and you will never get 1.

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Why is x(x+1) = 1? –  Anthony Sep 9 '13 at 23:14
    
@Tony $x^2+x+1=0\Leftrightarrow x(x+1)+1=0\Leftrightarrow x(x+1)=-1=1$. –  anon Sep 9 '13 at 23:24
    
@anon Why does -1=1? –  Anthony Sep 9 '13 at 23:26
    
@Tony Do you know what ${\Bbb Z}_2$ stands for here? –  anon Sep 9 '13 at 23:29
    
@anon It's just 0,1, right? –  Anthony Sep 9 '13 at 23:32

The operations your textbook refers to are just two abstract operations that are traditionally called “addition” and ”multiplication”, because they share many of the properties of addition and multiplication on rational or real numbers, as stated in the axiom for fields.

The set $\{a,b\}$ (where $a$ and $b$ are just two different objects) becomes a field under the operations \begin{gather} a+a=a,\quad a+b=b,\quad b+a=b,\quad b+b=a\\ aa=a,\quad ab=a,\quad ba=a,\quad bb=b \end{gather} Verifying the axioms for a field is just boring (and there are better ways than doing all the checks).

One might use different symbols and names for the operations, but this would only hinder computations, that instead are performed using the same rules of high school algebra.


Multiplication is not the same as addition, once you graduate from the natural numbers to rational numbers, where it can happen that $ab<a$ for positive $a$ and $b$, which of course is not possible in the natural numbers. It's not possible to consider $\frac{3}{2}\cdot\frac{1}{5}$ as a “repeated addition”.

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With regards to abstract fields, multiplication and addition refer to binary operations that satisfy certain axioms. They do not necessarily literally refer to a kind of multiplication or addition operation.

We sometimes use e.g. $2x$ as shorthand for $x+x$, but this is not multiplication. In fact, the $2$ might not even be defined in the field.

Wikipedia gives the "addition" and "multiplication" tables for the field of order $4$:

$$\begin{array}{c|cccc} + & 0 & 1 & A & B \\ \hline 0 & 0 & 1 & A & B \\ 1 & 1 & 0 & B & A \\ A & A & B & 0 & 1 \\ B & B & A & 1 & 0 \\ \end{array}$$

$$\begin{array}{c|cccc} × & 0 & 1 & A & B \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & A & B \\ A & 0 & A & B & 1 \\ B & 0 & B & 1 & A \\ \end{array}$$

If we repeatedly add $1$ to itself, we get the sequence $$1,0,1,0,1,\ldots.$$ But $A \times 1=A$, which cannot be obtained by repeated addition of $1$ with itself "$A$ times".

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How is $\pi \cdot e$ an addition? Or better yet, if $x$ is the variable in a polynomial ring, how is $g(x)\cdot f(x)$ an addition?

The "multiplication is addition" thing only works for integers.

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