Ring of polynomials in $\mathbb Z_{10}$ linear zero divisors I'm new here so apologies if I am not clear enough. I am trying to find the zero divisors of the form $ax + b$ in $\mathbb Z_{10}$. Specifically, I need to find the values of $b$. I know that $2,4,5,6$ and $8$ are zero divisors in $\mathbb Z_{10}$ but I am not sure how to translate these into linear divisors.
As all these zero divisors are the product of $5$ and itself (apart from $5$), can I just multiply $2x + b$ by $5$ to get $5(2x + b) = 5b \pmod{10}$? Then I would be left with $0,2,4,6$ and $8$.
I'm also looking to find the units in $\mathbb Z_{10}$ but I don't think there are any. I don't know how to show this, though.
Any help would be greatly appreciated.
Thanks, James
 A: If $ax + b$ is a zero divisor in $\mathbb Z_{10}[x]$, then $(ax+b)(cx+d)=0$, which implies $ac=0$, $ad+bc=0$, $bd=0$. In particular, $b$ is a zero divisor in $\mathbb Z_{10}$ because if $d=0$, then $c\ne0$ and $bc=0$.
Conversely, every $b$ that is a zero divisor in $\mathbb Z_{10}$ is a zero divisor in $\mathbb Z_{10}[x]$. 
A: Units
As you note, $0,2,4,5,6,8$ are zero divisors, hence cannot be units. We are left with $1,3,7,9$. $1$ is clearly a unit: it is its own inverse. Same holds for $9$: $9\cdot9=81\equiv1\pmod{10}$. As for $3$, note that $3\cdot7=21\equiv1\pmod{10}$. So $1,3,7,9$ are the units of $\mathbb{Z}_{10}$.
Linear zero divisors
As for the zero linear divisors, here are a few points.


*

*Note, as noted in the other answers, that if linear polynomial is a zero divisor, $(ax+b)(cx+d)=0$ for some $c,d\in\mathbb{Z}_{10}$.

*Now, the product expands to $acx^2+(bc+ad)x+bd$. So we must have:
$$ac=bc+ad=bd,$$
hence both $b$ and $a$ must be zero divisors.

*Now suppose $a,b$ are zero divisors, so there exist $c,d$ such that $ca=bd=0$. Is $bc+ad=0$? Not sure, but if we multiply it by $cd$ it will surely be. So if we take the polynomial $c^2dx+cd^2$, multiplied by $ax+bd$ it is 0.

*Of course, this opens another problem: is at least one of $c^2d,cd^2$ nonzero?

*If both $c$ and $d$ are even, their product will not be divisible by 10, hence $c^2d,cd^2$ will both be nonzero.

*If $c,d$ are both $5$, then $c^2d=d^2c=25\not\equiv0\pmod{10}$.

*If they are an even number and $5$, then the same will be true of $a,b$. So $ax+b$ will be either of the form $2kx+5$ or $5x+2k$, with $k$ an odd non-$5$ and nonzero number.

*Well, naturally I am assuming $a,b\neq0$, or at least one of them. If exactly one of them is nonzero, it means we have $2kx$ or $2k$, which is a zero divisor since multiplied by $5$ it gives 0$ $_, or we have $5x$ or $5$, which are zero divisors since $2\cdot5=10$_.

*What about the $2kx+5$ and $5x+2k$? For the first case, I must have $2kc=5d=2kd+5c=0$. For $2kc$ to be 0, $c$ must be $5$. For $5d=0$, $d$ must be $2m$. So $2kd+5c$ is $25+4km$. But that cannot be zero, since we would need $4km\equiv5\pmod{10}$, which is impossible since it is even. And I bet the case $5x+2k$ is dealt with in much the same way.
Bottom line theorem
$ax+b$ is a zero divisor if and only if one of the following holds:


*

*$a,b$ are both zero divisors of the same parity;

*$a=0$ and $b$ is a zero divisor;

*$a$ is a zero divisor and $b=0$.

