Prove that no linear polynomial in $\mathbb Z_{10}[x]$ is a unit I am trying to answer the following question about the ring of polynomials over $\mathbb Z_{10}[x]$:


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*Prove that no linear polynomial in $\mathbb Z_{10}[x]$ is a unit.


So far I have determined that:


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*$f(x)$ is a unit if $f(x)g(x)=1$.

*The units of $\mathbb Z_{10}[x]$ cannot be the zero divisors $(0,2,4,6,8)$ so we are left with $(1,3,7,9)$.

*$1$ is its own multiplicative inverse. Similarly, $9\cdot9\cdot9=81\equiv1(\mathrm{mod}10)$. Finally, $3\cdot7=21\equiv1(\mathrm{mod}10)$ so $(1,3,7,9)$ are in fact units.

*$\deg(f(x)g(x))=\deg(f(x))+\deg(g(x))$ so if $\deg(f(x))=1$ then $\deg(g(x))=-1$ (as $f(x)g(x)=1$ implies that $\deg(g(x))=-1$). The only unit in $\mathbb Z_{10}[x]$ whose multiplicative inverse is also an inverse is $1$. This implies $ax+b=1$.


This is where I get stuck. I have written “As $a$ and $b$ must be $(1,3,7,9)$ it is clear from inspection that $ax+b\equiv1(\mathrm{mod}10)$ is impossible” but I don't think this is a convincing argument as it doesn't even convince myself.
 A: Hint: One approach: If $(ax+b)p(x)=1$ in $\mathbb Z_{10}[x]$, then for any $c\in\mathbb Z_{10}$ $ac+b$ is a unit, since:
$$(ac+b)p(c)=1$$
Then show that if $a\neq 0$ you can always find a $c\in\mathbb Z_{10}$ such that $ac+b$ is not a unit.
If $a$ is a unit in $\mathbb Z_{10}$ then you can find $c$ so that $ac+b=0$.
If $a$ is divisible by $2$ but not $5$, find $c$ so that $ac+b$ is divisible by $5$.
If $a$ is divisible by $5$ but not $2$, find $c$ so that $ac+b$ is divisible by $2$.
A: We have this general result:

Let $R$ be a commutative ring with $1$. Then $f(X)=a_{0}+a_{1}X+a_{2}X^{2} + \cdots + a_{n}X^{n}$ is a unit in $R[X]$ if and only if $a_{0}$ is a unit in $R$ and $a_{1},a_{2},\dots,a_{n}$ are all nilpotent in $R$.

So, a linear polynomial $ax+b$ in $\mathbb Z_{10}[X]$ is a unit iff $b$ is a unit and $a$ is nilpotent in $\mathbb Z_{10}$. But since $10$ is squarefree, $0$ is the only nilpotent element. Thus, $a=0$ and $ax+b$ does not have degree $1$.
Here is a specialized proof of the general theorem for your case:
Suppose $(ax+b)(c_n x^n+ c_{n-1}x^{n-1} + \cdots + c_1 x + c_0)=1$. Then:
$a c_n=0$.
$a c_{n-1}+b c_n=0$, which implies $a^2 c_{n-1}=0$.
$a c_{n-2}+b c_{n-1}=0$, which implies $a^3 c_{n-2}=0$.
$\cdots$
$a c_{0}+b c_{1}=0$, which implies $a^{n+1}c_0=0$. Since $b c_0=1$, we get $a^{n+1}=0$.
A: HINT:
Let us show that if over a commutative ring $R$ we have
$$(a_0 + a_1 x)(b_0+ b_1 x + \cdots + b_n x^n) = 1$$
then $a_1^{n+1} = 0$. Indeed, from the above by $x \to \frac{1}{x}$ we get
$$(a_0 x + a_1 )(b_0 x^n + b_1 x^{n-1} + \cdots + b_n ) = x^{n+1}$$
Now, we must have $a_0 b_0=1$ so $a_0$ is invertible. So write 
$a_0 x + a_1 =a_0(x-\alpha)$. Now take the value $x = \alpha$ in the above equality and get $0 = \alpha^{n+1}$ and so $a_1^{n+1}=0$.
$\bf{Added:}$ In fact in a similar way we can show that if $f$ is of degree $m$ and $g$ of degree $n$ and $f g = 1$ then all the coefficients $a_1$, $\ldots$, $a_m$ are nilpotent. Indeed, by the same $x\to \frac{1}{x}$ we get 
$$(a_0 x^m + \cdots + a_m)(b_0 x^n + \cdots + b_n) = x^{m+n}$$ Now, since $a_0$ is invertible we can decompose $f$ in a larger ring $R'$
$$a_0 x^m + \cdots + a_m = a_0(x-\alpha_1)\cdots (x-\alpha_m)$$
so from the above, putting $x= \alpha_i$ in the above equality we get $\alpha_i^{m+n} = 0$ and from here we conclude that $a_1$, $\ldots $, $a_m$ are also nilpotent. For instance, $a_m^{m+n} = 0$. However, this is not optimal, as in fact $a_m^{1+n}=0$. 
