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I am trying to answer the following question about the ring of polynomials over $\mathbb Z_{10}[x]$:

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

So far I have determined that:

  • $f(x)$ 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.

gniourf_gniourf
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math_apprentice
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    Um, $3,7,9$ also have multiplicative inverses. It is not true that $deg(f(x)g(x))=deg(f(x))+deg(g(x))$. For example $(5x+1)(2x)=2x$ in $\mathbb Z_{10}[x]$. – Thomas Andrews Feb 17 '16 at 13:24
  • Thanks for your comment. What are the multiplicative inverses of $3,7,9$? It says in my notes that $deg(f(x)g(x))=deg(f(x))+deg(g(x))$ so why is this not true? I see your example but I don't know how you have obtained it? – math_apprentice Feb 17 '16 at 13:28
  • There are specific cases when $\deg(f(x)g(x))=\deg(f(x))+\deg(g(x))$. When $f,g$ are monic (have leading coefficient $1$) or when the coefficients are in a field ($\mathbb Z_{10}$ is not a field.) $3\cdot 7=1$ and $9\cdot 9=1$ in $\mathbb Z_{10}$. – Thomas Andrews Feb 17 '16 at 13:30
  • Sorry, I got confused there. I put in my original question that $3\cdot7=1$ and $9\cdot9\cdot9=1$. Okay, so it is only true in a field... Can you think of what the actual answer is as I seem to be barking up the wrong tree completely? – math_apprentice Feb 17 '16 at 13:34
  • I just wrote it out and see $10x^2=0$. Sorry. – math_apprentice Feb 17 '16 at 13:36

3 Answers3

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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 $2$.

Thomas Andrews
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  • Awesome. Thank you @Thomas Andrews. Much appreciated. – math_apprentice Feb 17 '16 at 13:45
  • I was hesitant to give this answer, because a common error in learning about polynomial rings is to confuse the polynomials with the functions they represent. For example, in $\mathbb Z_{p}$ the two functions $x$ and $x^p$ are equal, but in the ring $\mathbb Z_{p}[x]$, the polynomials are not equal. It does go the other way, however: Two polynomials that are equal are equal as functions (which is what I've used above.) – Thomas Andrews Feb 17 '16 at 13:55
  • Thanks again. It shows a gap in my understanding which I will discuss with my professor. – math_apprentice Feb 17 '16 at 14:01
  • jacobson radical here, neat – orangeskid Feb 17 '16 at 18:09
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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 $a^{n+1}=0$.

Community
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lhf
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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 $a_m^{1+n}=0$.

orangeskid
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