$(\Bbb Z[x]/(x^{n+1}))^\times\cong\Bbb Z/2\Bbb Z \times \prod _{i=1}^n \Bbb Z$ I'm trying to prove the group isomorphism $(\Bbb Z[x]/(x^{n+1}))^\times\cong\Bbb Z/2\Bbb Z \times \prod _{i=1}^n \Bbb Z$.
Obviously I tried to establish a ring isomorphism from $\Bbb Z[x]/(x^{n+1})$ to some ring $R$, a direct product of easier rings, and prove that $R^\times$ equals the RHS of the original isomorphism. In the solutions of similar problems I've seen, one defines a surjective ring homomorphism $\phi : R^\prime[x]\rightarrow R$ of substituting $x$ with an element of $R$ such that the kernel of $\phi$ equals the dividing ideal, and uses the fundamental homomorphism theorem and CRT to show the ring isomorphism.
However this doesn't work for this particular problem; since $x^{n+1}$ has a (unique) multiple root, the kernel of homomorphism of plugging in does not equal to the dividing ideal.  Moreover, since $\Bbb Z[x]$ is not an Euclidean domain, I have no idea what the units of $\Bbb Z[x]$ modulo some ideal are like.
I would appreciate your help.
 A: In any ring $R$, if $a \in R$ is nilpotent and $u\in R$ is a unit, then $a+u$ is a unit. Indeed, if $a^n=0$ then since $u^n = u^n - (-a)^n = (u+a)(u^{n-1} + u^{n-1}(-a)+\dots + (-a)^{n-1})$ is a unit, so is $u+a$. Nilpotent elements act on the units as infiniteseimal translations.
Equipped with this result, it is easy to see that the units of $R=\mathbf Z[x]/(x^{n+1})$ are all of the form 
$$\pm 1 + a_1x + \dots + a_nx^n$$
where $a_i \in \mathbf Z$. However we can't just map this to $(\pm 1, a_1, \dots, a_n)$, because the map is not a group homomorphism! What we need is the logarithm. Consider the map
$$\log: 1+x\mathbf Q[[x]] \to x \mathbf Q[[x]]\: : \:  1+xf \mapsto -\sum_{n=1}^\infty \frac {(-xf)^n}{n}.$$ 
Remark that the series actually converges in $\mathbf Q[[x]]$ with the $x$-adic topology. Moreover it is an additive map: $\log(1+xf) + \log (1+xg) = \log((1+xf)(1+xg))$ (this is an identity in power series). Now consider its restriction to $1+x\mathbf Z[x]$ and compose it with the canonical projection $x\mathbf Q[[x]] \to x\mathbf Q[x]/(x^{n+1})$. This induces an injective homomorphism of abelian group $$\log: 1+xR \to x\mathbf Q[x]/(x^{n+1}),$$
which identifies $1+xR$ with a lattice in the $\mathbf Q$-vector space $x\mathbf Q[x]/(x^{n+1})$ of dimension $n$. Therefore $1+xR$, which is obviously torsion-free, is a free abelian group of rank $\leq n$. Now, in order to show that it has rank $n$, we have to exhibit $n$ independent units. But $1+x, 1+x^2, \dots, 1+x^n$ are independent, because their images by $\log$ are independent in $x\mathbf Q[x]/(x^{n+1})$.
A: This problem is stated deceptively in the sense that it gives a highly polished description of what $\def\Z{\mathbb{Z}}(\Z[x]/(x^{n + 1}))^\times$ is, but doesn't actually say what it is as a subset of $\Z[x]/(x^{n + 1})$ itself.  You will have an easier time proving it if you sit down and compute what the units actually are, rather than trying to find an abstract homomorphism that happens to work.
First of all: the typical representation of $\Z[x]/(x^{n + 1})$ takes it to be the set of polynomials of degree $n$ or less, with the natural product reduced mod $x^{n + 1}$.  So among its elements are all integers $n \in \Z$, with their normal product since they all have degree zero, and the only units there are $\pm 1$.  That's your $\Z/2\Z$.
Second, you have to figure out what makes a polynomial $p(x)$ invertible mod $x^n$.  I claim these are just all the polynomials with constant term $\pm 1$; I leave it up to you to figure out how to find an inverse for such a thing.
Now, there is a set-theoretic isomorphism between $\{p(x) \mid \deg p \leq n, p(0) = \pm 1\}$ and $\Z/2\Z \times \prod_{i = 1}^n \Z$, namely, just take the coefficients of $p$, but this is not a group homomorphism because polynomial multiplication mixes the coefficients.  So there is the issue of representing this set in some other way than as a $\Z$-linear combination of powers of $x$ that makes it clearer what the multiplicative structure is.
The goal is to factor each polynomial in $\Z[x]/(x^{n + 1})$ into powers of $\pm 1$ and of $n$ other, predetermined polynomials.  The $\pm 1$ part is easy: it's $p(0)$.  For the next, consider the following trick:
$$(1 + kx + O(x^2))(1 - kx + O(x^2)) = 1 + O(x^2).$$
(This is seventh-grade algebra.)  Note that $(1 - x)^k = 1 - kx + O(x^2)$.  In other words, if we let $p_1(x) = 1 - x$, then every polynomial $p(x)$ has a unique factor of a power of $p_1(x)$ such that the quotient has no linear term.  Now, replace $x$ by $x^2$ and do the same with the quadratic term and $p_2(x) = 1 - x^2$, and so on up to $p_n(x) = 1 - x^n$.  (This is closely related to proving that $\Z[x]/(x^{n + 1})$ is what I said it was.)  If you work out the details, that is your proof.
