# Find members of this field.

Let $m$ be an integer, $m \geq 2$. and let ${\mathbb Z}_m$ be the set of all positive integers less than $m$, ${\mathbb Z}_m = \{0, 1, ..., m -1\}$. If $a$ and $b$ are in ${\mathbb Z}_m$, let $a + b$ be the least positive remainder obtained by dividing the (ordinary) sum of $a$ and $b$ by $m$, and, similarly, let $ab$ be the least positive remainder obtained by dividing the (ordinary) product of $a$ and $b$ by $m$. (Example: if $m = 12$, then $3 + 11 = 2$ and $3.11 = 9$.)

a) What is $-1$ in ${\mathbb Z}_5$?

b) What is $1/3$ in ${\mathbb Z}_7$?

I am unsure how $-1$ and $1/3$ can exist in the sets as it is meant to be made of positive integers and i know it forms a field when $m$ is a prime number, though i have n't been able to prove that yet either as i could not find justification for the following field properties.

• A(4) To every a there corresponds a unique scalar $-a$ such that $a + (-a) = 0$.
• B(2) Multiplication is associative $a(bg) =(ab)g$.
• B(4) To every non-zero scalar a there corresponds a unique scalar $a^{-1}$ (or $1/a$) such that $aa^{-1} = 1$.

Any help even just with the original two questions i think may give me enough to have another attempt at this proof.

Any help would be highly appreciated. Cheers

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To find -1 in $\mathbb{Z}_5$, recall that the symbol "-1" stands for the (unique) element which, when added to 1, yields zero. You only have five possibilities to check. (The problem of finding 1/3 in $\mathbb{Z}_7$ is similar.) –  Jon Jan 28 '12 at 23:18
If $1/3 \equiv a \bmod 7$, then $3a \equiv 1 \bmod 7$. Or, $3a = 1 + 7k$ for $k>0$. You can either construct the multiplication table for $\mathbb{Z}_7$ and look for such number $a$ such that $3a = 1$ in the table, or use the Euclidean algorithm. –  user2468 Jan 28 '12 at 23:24
@Norbert $3 \cdot 9 = 27$. It might be easier to note that $3 \cdot 5 = 15$. –  Dylan Moreland Jan 28 '12 at 23:26

Defining $\mathbb Z_m$ as you did (the set of elements... + operations...), sometimes we abuse of notation. For example, we write $-1$ but we already are refering to the least positive remainder obtained by dividing $-1$ by $m$. For example, $-1$ in $\mathbb Z_{12}$ means $11$ since $$-1= 12\times 0 + 11.$$ Actually, this happens with any integer number. $127$ is $7$ in $\mathbb Z_{12}$ since $$127=12\times 10 + 7.$$

With respect to the quotients, some elements have an inverse element. In you example b), $3$ has a inverse in $\mathbb Z_7$ since $$3\times 5= 15 =7\times 2+1,$$ thus we write $3\times 5=1$ and we say that $5$ is the inverse of $3$, and sometimes we write $5=1/3.$ Note that this does not always happen. For example, $3$ has not an inverse element in $\mathbb Z_{12}$. Actually, $k$ has an inverse element in $\mathbb Z_{m}$ if and only $k$ and $m$ are coprime. One consecuence is that $\mathbb Z_m$ is a field if and only if $m$ is a prime number.

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In a field $F$, if $a\in F$, then "$-a$" represents the unique element $x$ of $F$ with the property that $a+x = 0$.

In a field $F$, if $a\in F$, $a\neq 0$, then $\frac{1}{a}$ (or $a^{-1}$) represents the unique element $y$ of $F$ with the property that $ay = 1$.

So: to find $-1$ in $\mathbb{Z}_5$, you need to find the unique element $x$ of $\mathbb{Z}_5$ that, added to $1$, yields $0$ (that is, the unique solution to $x+1=0$ in $\mathbb{Z}_5$). And to find the element $\frac{1}{3}$ of $\mathbb{Z}_7$, you need to find the unique element $y$ of $\mathbb{Z}_7$ that, when multiplied by $3$, yields $1$. That is, the unique solution to $3y=1$ in $\mathbb{Z}_7$.

The reason why in a field $F$, given $a\in F$ there is a unique $x$ such that $a+x=0$ is that the field is supposed to be an abelian group under addition. This means $+$ must be associative and commutative, that F has to have an element $0$ such that $a+0 = a$ for all $a$; and that for all $a$ there must exist a $b$ such that $a+b=0$; the fact that these are unique follow easily from these properties.

The reason why in a field $F$, given $a\in F$, $a\neq 0$, there must exist a unique $y$ such that $ay=1$ is again part of the definition: a field is required to have a multiplicative identity that is not the additive identity (an element $1\neq 0$ such that $a1 = a$ for all $a$), and every nonzero element is required to have a multiplicative inverse; uniqueness follows from these facts. In short: the reason A(4), B(2), and B(4) must hold is by definition. There is no further justification needed: in order to be able to be called a "field", those properties (among others) must be satisfied. It is a membership requirement to be able to get past the bouncer at the door of the Club of Fields.

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Thanks guys, I guess for part a) the answers would 4, 9, 14, 19 ...for part b) the answer would 7, 14,.. –  Hardy Jan 28 '12 at 23:57
@Hardy: No. Look at your definition of $\mathbb{Z}_m$. Does $\mathbb{Z}_5$ include $9$, $14$, $19$, or any of the others? Does $\mathbb{Z}_7$ include $7$, $14$, etc? And is $3\times 7$ really congruent to $1$ modulo $7$? –  Arturo Magidin Jan 29 '12 at 0:06
thanks i think - 1 in Z5 is 4 and 1/3 in Z7 is 5. –  Hardy Jan 29 '12 at 0:48
@Hardy: You think correctly. –  Arturo Magidin Jan 29 '12 at 0:58

$$3\cdot 5 \text{ is the same as }1\text{ in }\mathbb{Z}_7.$$ This should tell you how $1/3$ can exist in $\mathbb{Z}_7$.

$$(1 + \text{something}) \text{ is the same as } 0\text{ in }\mathbb{Z}_7.$$ This should tell you how $-1$ can exist in $\mathbb{Z}_7$.

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