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Prove that the polynomial $P=2X+1 \in \Bbb Z_4[X]$ has an inverse element. What happens if we consider $P$ as an element of $\Bbb Q [X]$?

If $P \in \Bbb Z_4[X]$, then any $Q=kX+1$ where $k \equiv 2 \pmod{4}$ will work as an inverse right? I suppose that the result will not change in $\Bbb Q [X]$ as the coefficients have a multiplicative inverses except $0$?

mrtaurho
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Rico Jello
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  • You write 'any' $Q$, but your description specifies a single polynomial in $\mathbb{Z}_4[X]$ namely, $2X + 1$. It does solve your first question. – Krijn Dec 13 '21 at 21:36
  • Woulnd't something like $6X+1$ work? – Rico Jello Dec 13 '21 at 22:03
  • For the first questions $P$ is a unit+nilpotent which is trivially invertible - see the first linked dupe. For the second look at the lead coef of $PQ = 1$ (more generally over any commutative ring see the 2nd linked dupe). – Bill Dubuque Dec 14 '21 at 10:19

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You are correct regarding $Q=kX+1$ being an inverse whenever $k\equiv2\mod 4$.

However, the result changes drastically if we move to $\mathbb Q[X]$ as this is a polynomial ring of an integral domain. You should convince yourself that for $R$ and integral domain and $f,g\in R[X]$ we have

$$ \deg fg=\deg f+\deg g $$

making invertibility for any non-constant polynomial impossible.

mrtaurho
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  • How does this make the invertibility impossible? – Rico Jello Dec 13 '21 at 22:08
  • @RicoJello What's the degree of the constant polynomial $1$? :) – mrtaurho Dec 13 '21 at 22:21
  • It's zero and the degree of $P$ is one. We would have to have that $\deg(PQ)= 0 = \deg(P) + \deg(Q) = 1 + \deg(Q)$, but this will never be true. I still feel like they've would have wanted a single $Q$ for the first part. It seems that $Q= kX +1$, when $k \equiv 2 \pmod{4}$ is the same as $P$? – Rico Jello Dec 13 '21 at 22:27
  • @RicoJello Both of what you just said is true. – mrtaurho Dec 13 '21 at 22:51
  • @Rico $\,(1+nx)(1-nx)\equiv 1\pmod{n^2}.\,$ When $n=2$ we have $\,-2\equiv 2\,$ so $\,1 \equiv (1+2x)(1-2x) = (1+2x)^2.\,$ That's why $Q=P$ here, i.e. because $\,-2\equiv 2\pmod{4}.\ $ See unit+nilpotent inversion (and the simpler multiple method) in the first linked dupe for further intuition. – Bill Dubuque Dec 14 '21 at 10:28