Field definition - Unique inverse

Question

In Abstract Algebra: Theory and Applications, every non-zero element in a field has an unique multiplicative inverse. However, on Wikipedia the definition has dropped the unique term.

For example, is $\mathbb{Z}_{18}$ a field? For example, $ 3 \cdot 6 = 3 \cdot 12 = 0 \ \text{mod 18}$ so 3 has atleast two multiplicative inverses.

Answer 1

In a field $F$, a multiplicative inverse of an element $a$ is an element $b$ such that $ab=ba=1$. If the axioms that you are working with tell you that every element $a\ne0$ has a multiplicative inverse, then there is no need to add that such inverse is unique. In fact, if both $b$ and $c$ are multiplicative inverses of $a$, then\begin{align}b&=b.1\\&=b.(a.c)\\&=(b.a).c\\&=1.c\\&=c.\end{align}And $\Bbb Z_{18}$ is not a field because, for instance, $3$ has no multiplicative inverse.

Answer 2

In a field $F$, the non-zero elements form a group with respect to multiplication. Thus if a multiplicative inverse exist , it must be unique.

It is easy to show that any integral domain must have characteristic either $0$ or a prime $p$.

i.e. $(1+1+...+1)=0\implies$ if $char F= mn\,,\,m,n<char F$ then $(m\cdot 1)(n\cdot 1)=0$ so either $m\cdot 1=0$ or $n\cdot 1=0$ . That is $char F=m$ or $n$. So $char F$ is not composite. i.e. it is a prime.

Now for any finite field $F$ of characteristic $p$, you have an surjective homomorphism from $f:\Bbb{Z}\to F$ given by $f(n)=n\cdot 1$.

Then the kernel is $(p\Bbb{Z})$ and hence any finite field contains $\frac{\Bbb{Z}}{(p\Bbb{Z})}=\mathbb{Z}_{p}$.

Then $F$ forms a vector space over $ \mathbb{Z}_{p}$ and it must be finite dimensional as $F$ is finite.

So say it has dimension $n$ with basis $\{a_{1},a_{2},...,a_{n}\}$

Then $$F=\{\sum_{i=1}^{n}c_{i}a_{i}:c_{i}\in \mathbb{Z}_{p}\}$$

So there are $p$ choices for each $c_{i}$ and hence $|F|=p^{n}$.

So any finite field is of cardinality $p^{n}$ for some prime $p$ and positive integer $n$.

Answer 3

Well, the residue class ring $\Bbb Z_n$ is a field if and only if $n$ is a prime.

Uniqueness is established as follows:

Let $b,c$ inverses of $a\ne 0$. Then $b = b\cdot 1 = b \cdot (a\cdot c) = (b\cdot a)\cdot c = 1\cdot c = c$.

Answer 4

You're confusing about the definition of inverse. An inverse of an element $a$ is another element $b$ such that $ab=1$ (and $ba=1$, but in this case there's no need to specify this).

The fact that $3\cdot6=0$ in $\mathbb{Z}/18\mathbb{Z}$ implies that $3$ has no inverse: if there were one, say $b$, we'd have $1=3b$ and therefore $$ 6=6\cdot1=6\cdot3b=0b=0 $$ An element $a$ such that there exists $b\ne0$ with $ab=0$ is called a zero-divisor (sometimes there is the clause that $a\ne0$). In rings of the form $\mathbb{Z}/n\mathbb{Z}$, every element is either invertible or a zero-divisor (this however doesn't happen generally in commutative rings).

Uniqueness of the inverse is valid for every operation having a neutral element. Suppose you have a monoid $X,\cdot$ with neutral element $1$. An inverse of $a\in X$ is an element $b\in X$ such that $ab=1=ba$ (the operation need not be commutative).

It's general theory that when the operation is associative, an inverse is unique when it exists: if $ab=1=ba$ and $ac=1=ca$, then $$ b=b1=b(ac)=(ba)c=1c=c $$ No ring or field involved, just an associative operation with neutral element.

You're referring to section 16.1 of the textbook you link. Well, when the definition of a group the author states the condition

for each element $a\in G$ there exists an inverse element in $G$, denoted by $a^{-1}$, such that $a^{-1}\circ a=a\circ a^{-1}=e$.

I find this a bad definition, because there is no mention that such an element must be unique, so we can use the notation $a^{-1}$ to begin with. However, proposition 3.18 states uniqueness of the inverse (without proof). I'm not sure why the authors change their mind when dealing with rings (I suspect that the two parts were written by different authors).