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I came across the question:

Determine if the element $x^3 + 2 + \langle x^4 − x^2\rangle\in (\mathbb{Z}/5[x])/\langle x^4 − x^2\rangle$ is a unit.

Notation: $\langle x^4 − x^2\rangle$ is the ideal generated by $ x^4 − x^2$.

My guess is that it is a unit, but I can't figure out how to calculate the multiplicative inverse. I know that for it to be a unit, there has to be a polynomial $p(x)\in (\mathbb{Z}/5)[x]$ such that \begin{equation} p(x)(x^3 +2)-1\in \langle x^4-x^2\rangle . \end{equation} In other words, there has to be polynomials $p(x)$ and $q(x)$ such that \begin{equation} p(x)(x^3 +2)= q(x)(x^4-x^2)+1. \end{equation} Which implies $p(x)$ has to have a constant term that is equal to $3$. But I don't know how to proceed from this point.

In an attempt to use another approach, I used the division algorithm which lead me to \begin{equation} (x^2+3x+1)(x^4-x^2)= (x^3+3x)(x^3+2). \end{equation} And this made me rethink, I am actually not sure if $x^3 + 2 + \langle x^4 − x^2\rangle$ is a unit now.

Any help is gladly appreciated.

P3p3O
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  • Related. You can use the extended Euclidean algorithm to find the inverse. Examples of that (there are many on this site) typically have an irreducible polynomial in place of $x^4-x^2=x^2(x-1)(x+1)$, but the algorithm works the same as long as the element to be inverted is represented by a coprime polynomial. – Jyrki Lahtonen Dec 15 '24 at 05:36
  • By here in the linked dupe: $,f,$ is a unit in $,\Bbb F_5[x]/(g)\iff \gcd(f,g)=1,$ in $,\Bbb F_5[x]\ \ $ – Bill Dubuque Dec 15 '24 at 05:51
  • Note that the coprimality $,\gcd(f,g)=1,$ can be quickly verified w/o the Euclidean algorithm since here $f$ is coprime to $,x,,x\pm1,$ by $,f(0),f(\pm1)\neq 0$ in $\Bbb Z_5,,$ hence $f$ is coprime to their product $, g = x^2(x-1)(x+1),,$ i.e. units (invertibles) in the quotient are closed under product. $\ \ $ – Bill Dubuque Dec 15 '24 at 06:06

2 Answers2

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In $\mathbf Z/m\mathbf Z$, $a$ is a unit if and only if $a$ and $m$ are relatively prime.

Similarly, when $K$ is a field and $f(x) \in K[x]$, $f(x)$ is a unit in $K[x]/(g(x))$ if and only if $f(x)$ and $g(x)$ are relatively prime.

You ask abou $K = \mathbf Z/5\mathbf Z$, $f(x) = x^3+2$, and $g(x) = x^4-x^2 = x^2(x-1)(x+1)$. Do you see any nonconstant common factors between $f(x)$ and $g(x)$ in $(\mathbf Z/5\mathbf Z)[x]$?

KCd
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  • No, there are no common factors. So it is a unit! But now I have two questions: 1) How could one go about proving the first assertion (with $f$ and $g$ relatively prime). 2) is it still possible to find an inverse? – P3p3O Dec 15 '24 at 05:04
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    If you know how to prove in $\mathbf Z$ that $a \bmod m$ is invertible if and only if $a$ and $m$ are relatively prime, then adapt that argument to $K[x]$. If you do not know how to prove the result in $\mathbf Z$, then learn that first. Similarly, to invert $f(x) \bmod g(x)$, take a method that computes the inverse of $a \bmod m$ and adapt that to the polynomial setting. In short, make sure you can treat the more basic case of $\mathbf Z$ and then adjust the method there to handle polynomials. – KCd Dec 15 '24 at 05:19
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. – Bill Dubuque Dec 15 '24 at 05:51
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Just like to compute modular inverses, you can do an extended GCD algorithm to get $af + bg = \gcd(f, g)$ and if that gcd is $1$ (or another unit) then $bg \equiv 1 \pmod f$.

Here we have (quotients and remainders, using long division):

\begin{align} x^4 + 4x^2 &= x(x^3 + 2) + (4x^2 + 3x) \tag1 \\ x^3 + 2 &= (4x + 2)(4x^2 + 3x) + 4x + 2 \tag2 \\ 4x^2 + 3x &= (x + 4)(4x + 2) + 2 \tag3 \end{align}

So we get $2$ (a unit) as our gcd. Then we can solve for $a(x)$ and $b(x)$ by solving each of these quotient/remainder equations for the remainder and substituting each into the next.

For example, equation (1) says that

$$f = xg + (4x^2 + 3x) \longrightarrow4x^2 +3x = f + 4xg$$

and then you can substitute that into (2):

$$g = (4x + 2)(4x^2 + 3x) + 4x + 2$$

to get $4x + 2 = (\cdots) f + (\cdots) g$. And then substitute that into (3).

I also recommend at some point starting to use a computer algebra system to do these calculations. E.g. in SageMath it's

R.<x> = GF(5)[]
f = x^4 - x^2
g = x^3 + 2
xgcd(f, g)

ee https://doc.sagemath.org/html/en/tutorial/tour_polynomial.html for info about the RingName.<variables> = BaseRing[] notation.

Sera Gunn
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