I've been asked if given $R=\Bbb Z/6\Bbb Z$, to come up with nonzero polynomial $f$ in $R[x]$ of degree at most $3$ such that each element of $R$ is a root of $f$.
My question is, am I allowed to let $f=sin(\bar2 x \pi)$ or am I not allowed to come use trig functions or $\pi$ based on how the polynomial ring is defined? If I am not allowed to use use trig, can I get some help on this?
Thanks!
Your $f(x)$ isn't a polynomial.
Hint: $6$ divides any product of three consecutive integers.
Polynomials are finite linear combinations of $1,x,x^2,\dots$, so trig polynomials are not allowed.
Now, if $0$ is a root, we have that $f(x) = x(x-a)(x-b)$. Now, let $a = 1$ just to see what happens. So: $$f(x) = x(x-1)(x-b)$$ Let's look at what $f(2),f(3),f(4),f(5)$ are :
$$f(2) = 2(2-b),\quad f(3) = 3\times 2\times (3-b),\quad f(4) = 4\times 3\times (4-b),\quad f(5) = 5\times 4\times (5-b)$$
Now, recall that $6 \equiv 0\pmod{6}$, so $f(3) = 0$, and $f(4) = 2\times 2\times 3 \times (4-b) = 0$.
So, we just need $f(2)$ and $f(5) = 0$. So, we need $(2-b) = 0,3$, and $(5-b) = 3,0$. Choosing $b = 2$ seems to work.
So, we get that: $$f(x) = x(x-1)(x-2)$$ is of degree $3$, and zero for all $x\in \mathbb{Z}/6\mathbb{Z}$.
Observe that
$3 \cdot (\text{"even" number})=0$ in $\Bbb Z_6$.
Either $x$ or $x+1$ is even in $\Bbb Z$.
From above two observations, $3x(x+1)$ serves the purpose.
Note: I've put the word 'even' in double quotes since in the ring $\Bbb Z_6$ there is nothing even or odd. The 'even' I used is from $\Bbb Z$. Also the first observation means that, an even multiple of 3 is a multiple of 6 in $\Bbb Z$.
You can also go for a direct approach. Suppose that $f(x)=ax^3+bx^2+cx+d$ is such a polynomial, then \begin{eqnarray} d&=&0\\ a+b+c+d&=&0\\ 2a+4b+2c+d&=&0\\ 3a+3b+3c+d&=&0\\ 4a+4b+4c+d&=&0\\ 5a+b+5c+d&=&0.\\ \end{eqnarray} Now find a non-zero solution to his system of equations in $\mathbb{Z}_6$.
