Solve an equation with $\mathbb{Z}_3$

Question

I can't solve the following equation system using $\mathbb{Z}_3$: \begin{align} &x+y=1,\\ &2x+y=0. \end{align} $\mathbb{Z}_3$ is a field with the numbers $(0,1,2)$ and modular operations.

What I tried: I added $2$ to the first equation and got:

$$ x+y+2= 0 = 2x + y,$$

$$x+2=2x.$$

And I don't know what to do from here, i know it seems obvious $x=2$ but we just learned all the $11$ axioms of a field, and I can only use them or the $5$ properties that derived from them. My problem is that I don't know how to express the $2x$ in a way that can help solve it.

Thanks for help.

Answer 1

You can solve it in exactly the same way as you would in the rational or real numbers; solve for $y$ the first equation, $y=1-x$, and substitute in the second one: $$ 2x+(1-x)=0 $$ that becomes $$ x=-1 $$ or $x=2$, since you are in $\mathbb{Z}/3\mathbb{Z}$. Now you get $y=1-(-1)=2$.

Or sum the two equations, getting $x+2x+y+y=1$, so $2y=1$. Since $2^{-1}=2$, because $2\cdot 2=1$, you have again $y=2$. Subtracting the first equation from the second gives $x=-1$, that is, $x=2$.

In general you can compute inverses in fields of the form $\mathbb{Z}/p\mathbb{Z}$ ($p$ a prime) using the Euclidean algorithm and Bézout's theorem.

Answer 2

Since $\frac{\mathbb{Z}}{3\mathbb{Z}}$ is a field, we know that $ab=ba$ for every $a,b \in \frac{\mathbb{Z}}{3\mathbb{Z}}$ and that this product is zero if and only if one of $a$ or $b$ is zero. So, we know that $2x=2(2)$ cannot be zero in this field. Now we just need to deduce what it is! This is the same as observing that for every nonzero $a \in \frac{\mathbb{Z}}{3\mathbb{Z}}$ there is a unique element $b$ so that $ab=1=ba$ (in larger fields you would have a harder time and would actually need to pull out the largest multiple of $p$, where your field is of the form $\frac{\mathbb{Z}}{p\mathbb{Z}}$, but that is equivalent for this field). So, if we look at the nonzero elements, we have the set $\lbrace 1,2 \rbrace$. Well, we know that for any number $a$, $1a=a=a1$, so it remains for you to go through and check the results with $a=2$ in order to find what the inverse of $a$ is (note that the inverse of $a$ is in the set $\lbrace 1,2 \rbrace$). This will tell you immediately what $2x=2(2)$ is in $\frac{\mathbb{Z}}{3\mathbb{Z}}$.

Answer 3

Ok, you must work $(\mod 3).$ So the first equation is written $y=1-x=1+2x$ in ${\bf Z}_3.$ Now the second equation is written as $2x+(1+2x)=0$ thus $4x+1=0$ or $x+1=0$ or $x=-1=2.$ So $y=1+2\cdot 2=1+4=5=2.$ The desire solution is $(x,y)=(2,2).$