Are there any easier and generalized methods for finding group of units (in quotient of polynomial ring)?

Question

Here are some problems that I've stuck.

Find the each group of units for given 2 quotient rings.

1) $\Bbb Z_4[x]/\langle x^3\rangle $

2) $\Bbb Z_7[x]/\langle x^2-x\rangle $

In the case of example 1, I tried those like putting the quotient's element form $\{ax^2+ bx +c\}$. It is trivial that inverse form also like that.

Hence, all we have to do is just find the

$\{dx^2+ ex +f \}$ s.t. $(ax^2+ bx +c)(dx^2+ ex +f)\in 1+\langle x^3\rangle $

BUT I FAILED.

It seems like really having a complicated process when putting the element form for finding the inverse.

So are there any more simple and generalized methods for finding the inverse or group of units of the quotient of the polynomial ring ?

P.s.) Thanks to lisyarus, I post my process.

Answer 1

Both of these rings are finite, commutative and unital, and in any such ring, any element is either a unit or a zero divisor.

Proof: Take an element $r$ in such a ring $R$ and consider the ideal $\langle r\rangle$. This ideal is the image of the function $\mu_r:R\to R$ given by $\mu_r(x)= rx$.

If the ideal is the entire ring, then the image of $\mu_r$ contains $1$, meaning that $r$ is a unit. If the ideal is not the entire ring, then $\mu_r$ is not surjective and therefore injective. Because $\mu_r$ is a homomorphism from the additive group on $R$ to itself, non-injective means it must have a non-trivial kernel, showing that $r$ is a zero divisor. $\square$

This means that we can go look for zero divisors instead. They are, in my opinion, easier to spot in these rings. Whether the resulting algebra to confirm everything is actually easier, that's a different question, and maybe the answer to that is no.

Any (non-zero) element of $\Bbb Z_4[x]/\langle x^3\rangle$ with an even constant term is a zero divisor, as it may be multiplied by $2x^2$ to give zero. On the other hand, an element with an odd constant term cannot be a zero divisor. We see this my noting that whatever non-zero element we multiply it with will have a least-degree term. This least degree term won't vanish in the product.

$\Bbb Z_7[x]/\langle x^2-x\rangle$ isn't quite as straight-forward, but we can immediately see that $x$ and $x-1$ are zero divisors, as $x(x-1) = x^2-x$. I claim that these two, and any non-zero constant multiple of them, are the only zero divisors we have.

Indeed, let $(ax + b)(cx + d) = 0$ with neither factor being $0$. Then $$ 0 = (ax + b)(cx + d) = acx^2 + (ad+bc)x + bd = (ad+ac+bc)x + bd $$ From this we conclude that either $b$ or $d$ is $0$. Assume $b = 0$. Then we get $a(d+c) = 0$, and $a$ cannot be $0$ since $ax+b\neq 0$. This means $d = -c$, and we have $ax\cdot c(x-1)$. Assuming $d = 0$ gives an analoguous result.

This shows that the only zero divisors in $\Bbb Z_7[x]/\langle x^2-x\rangle$ are multiples of $x$ and multiples of $x-1$. The remaining $36$ non-zero elements are units.

Answer 2

Taking an approach different from what Arthur suggests, I'll try to show slightly easier and generalized methods.

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1. $\mathbb Z_4[x]/\langle x^3\rangle$

First, lets go on with your approach. You already got to $(ax^2+bx+c)(dx^2+ex+f)=1$ implies $cf=1$, so $c$ has to be an invertible element of $\mathbb Z_4$, so $c=1$ or $c=3$, and $f=c^{-1}$.

Now, from $bf+ce=0$ we get $bc^{-1}+ce=0 \Rightarrow bc^{-2}+e=0 \Rightarrow e=-bc^{-2}$.

Finally, from $af+be+cd=0$ we get $ac^{-1}-b^2c^{-2}+cd=0\Rightarrow d=-ac^{-2}+b^2c^{-3}$.

So, given that $c$ is invertible, we have the formulas for $d,e,f$ in terms of $a,b,c$. Thus, $c$ being invertible is the necessary and sufficient condition for $ax^2+bx+c$ to be invertible.

You can see we never actually used any information about the ring of scalars $\mathbb Z_4$. This is no coincidence; $p \in R[x]/\langle x^n \rangle$ is invertible iff its constant term is invertible.

To give a different proof of this, observe that we already know that an invertible constant term is a necessary condition. Thus, $p = c + x\cdot q$ with $c$ invertible. Since $x^n=0$ in $R[x]/\langle x^n \rangle$, clearly $(xq)^n = x^nq^n=0$, so $xq$ is nilpotent. Now, in any commutative ring a unit plus a nilpotent is still a unit. Hence, if $c$ is invertible, $p$ is invertible.

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2. $\mathbb Z_7/\langle x^2-x\rangle$

This case is a bit easier, since $x^2-x = x(x-1)$, and $x$ and $x-1$ are coprime. By Chinese remainder theorem, we get

$$\mathbb Z_7/\langle x^2-x\rangle \cong \mathbb Z_7/\langle x\rangle \times \mathbb Z_7/\langle x-1\rangle \cong \mathbb Z_7 \times \mathbb Z_7$$

via the isomorphism

$$p(x) + \langle x^2+x \rangle \mapsto (p(0), p(1))$$

An element $(a,b) \in \mathbb Z_7 \times \mathbb Z_7$ is invertible iff $a \neq 0$ and $b \neq 0$, which translates to $p \in \mathbb Z_7/\langle x^2-x\rangle$ being invertible iff $p(0) \neq 0$ and $p(1) \neq 0$.

Let $p(x) = ax+b$. $p(0) \neq 0$ means $b \neq 0$, while $p(1) \neq 0$ means $a \neq -b$. These are the conditions for $p$ to be invertible.