# Thread: set of symbols form a field

1. ## set of symbols form a field

Hi all, I am going through a Algebra textbook and I need a bit of help with this question, so thank you in advance for any help.

Question: Prove that the set of symbols $\{a+bi \ | \ a,b \in F_3 \}$ forms a field with nine elements, if the laws of composition are made to mimic addition and multiplication of complex numbers. Will the same method work for $F_5$? For $F_7$? Explain.

So I can see that the nine elements are: $\{0,1,2,i,1+i,2+i,2i,1+2i,2+2i \}$ but I am confused as to where to go from here.

June

2. ## Re: set of symbols form a field

Well, write all the conditions for the set $\mathcal{S}_p$ of symbols to form a field. For example, taking into account that $F_p=\{0,1,\ldots, p-1\}$ with $p$ prime is a field, then. given $a+ib\in\mathcal{S}_p$ then,

$(a+ib)+(-a+(-b)i)=0+0i$

etc, etc,... . Show your work and we check it.

3. ## Re: set of symbols form a field

Thanks FernandoRevilla, sorry for my late reply, been gone a few days.

So,
Addition and Multiplication:
$(a+bi)+(c+di)=(a+c)+i(b+d)$ and $(a+bi)(c+di)=(ac-bd)+i(ad+bc)$

Inverse:
$(a+bi)^{-1}=\frac{a-bi}{a^2+b^2}$

Distributive:
$(a+bi)[(c+di)+(e+fi)]=(a+bi)(c+di)+(a+bi)(e+fi)$

The complex addition identity is 0.
The complex multiplication identity is 1.

And commutative and associative follow easily.

Is this correct working out?

Thanks

4. ## Re: set of symbols form a field

And what about $F_5$ and $F_7$, will the same method work for these?

Thanks

5. ## Re: set of symbols form a field

So I need to look at $\{a+bi \ | \ a,b \in F_5 \}$ and $\{a+bi \ | \ a,b \in F_7 \}$ to see if they form a field, if the laws of composition are made to mimic addition and multiplication of complex numbers.

So for $\{a+bi \ | \ a,b \in F_5 \}$ the elements are:
$\{0,1,2,3,4,i,2i,3i,4i,1+i,2+i,3+i,4+i,1+2i,1+3i,1 +4i,2+2i,2+3i,2+4i,3+2i,3+3i,3+4i,4+2i,4+3i,4+4i\}$

So where do I go from here?

Any help guys?

Thanks

6. ## Re: set of symbols form a field

Originally Posted by Juneu436
Inverse: $(a+bi)^{-1}=\frac{a-bi}{a^2+b^2}$
Right, but you need to verify that if $a+bi \neq 0$ then, $a^2+b^2\neq 0$ . Easily proved, this condition is verified in $F_3$ but not in $F_5$ : choose for example $1+2i$ ( $1^2+2^2=0$ ) . Try now with $F_7$ .