Let $p$ be prime and $n = p^2$. Show that the polynomial $1+px$ has a multiplicative inverse in $\mathbb{Z}_n[x]$. Is this polynomial invertible if $n = p^k$ for arbitrary $k>0$?
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4What do you denote $\Bbb Z(x)$, exactly? – Bernard Jan 22 '20 at 23:07

Sorry, should be Z_n(x) (field of integers mod n) – kt046172 Jan 22 '20 at 23:11

2If $n=p^2$, integers modulo $n$ do not make up a field: it even has zero divisors. – Bernard Jan 22 '20 at 23:13

a ring perhaps? Sorry, I am confused by the notation as well – kt046172 Jan 22 '20 at 23:14

6Maybe you mean $\mathbb{Z}_n[x]$ instead of $\mathbb{Z}_n(x)$? – Daniel Schepler Jan 22 '20 at 23:14

It is indeed a local ring with a nilpotent maximal ideal. So afield of rational fractions with coefficients in this ring does not exist, only the ring of polynomials. – Bernard Jan 22 '20 at 23:23
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$(1+px)(1px)=1p^2x^2=1$ and so $1+px$ has a multiplicative inverse.
If $n=p^k$ the same idea works. The inverse is $$1px+(px)^2(px)^3+...$$
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Well we need $(1+px)P(x) \equiv 1$ so if $P(x) = \sum_{k=0}^m a_k x^k$ we must have
$(1+px)\sum_{k=0}^m a_k x^k = \sum_{k=0}^m a_k x^k + \sum_{k=0}^m pa_k x^{k+1}= a_0 + \sum_{k=1}^m (pa_{k1}+ a_k)x^k + pa_m x^{m+1}$ were
$a_0 \equiv 1 \pmod n$ for $k=1,....m$ that $pa_{k1}\equiv a_{k}\pmod n$ and that $pa_m \equiv 0 \mod n$
So $pa_0 \equiv p \equiv a_1\pmod n$ so $a_1 \equiv p\pmod n$. And $p*a_1 = p^2\equiv 0\equiv a_2$ so $a_2 \equiv 0\pmod n$.
So $P(x) = 1  px$ and $(1 + px)(1px) = 1p^2x^2 \equiv 1\pmod n$.
That's it.
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