Is $A\prod A$ isomorphic to $A[t]/(t^2-1)$? Let $A$ be a ring, given two ring structure on abelian group $A\oplus A$ by $(a,b)\times(c,d)=(ab,cd)$ and $(a,b)\times(c,d)=(ac+bd,ad+bc)$, in other words, view it as $A[t]/(t^2-1)$. Are these two rings isomorphic? 
More generally, if $A$ is a ring, $A[t]/(t^n-1)$ has a primitive root of unity, is $A[t]/(t^n-1)$ isomorphic to $\prod_nA$?
 A: Let me assume that when $A$ is noncommutative then by $A[t]$ you mean $A$ with a central element $t$ adjoined. There is a natural map
$$A[t]/(t^2 - 1) \ni f(t) \mapsto (f(1), f(-1)) \in A \times A$$
and if $2$ is invertible in $A$ then this map is an isomorphism of rings, with inverse
$$(a, b) \ni A \times A \mapsto \frac{a+b}{2} + \frac{a-b}{2} t \in A[t]/(t^2 - 1).$$
On the other hand, if $A$ has characteristic $2$ then $A[t]/(t^2 - 1) \cong A[t]/(t - 1)^2$ and this need not be isomorphic to $A \times A$, for example if $A$ itself has no nontrivial nilpotents.
Similarly, if $A$ has all $n^{th}$ roots of unity $1, \zeta_n, \dots$ (Edit: We also need to require that these are central) then there is a natural map
$$A[t]/(t^n - 1) \ni f(t) \mapsto (f(1), f(\zeta_n), \dots) \in A^n$$
and if $n$ is invertible in $A$ then this map is an isomorphism of rings. The inverse is essentially given by the inverse discrete Fourier transform. 
As above this need not be true if $A$ has characteristic dividing $n$, but even if $A$ is, say, a field of characteristic $0$ there is a further obstruction coming from not having enough roots of unity. Over a field $k$ the chinese remainder theorem gives
$$k[t]/(t^n - 1) \cong \prod_i k[t]/f_i(t)^{m_i}$$
where $t^n - 1 = \prod_i f_i(t)^{m_i}$ is the prime factorization of $t^n - 1$ in $k[t]$. For example, when $k = \mathbb{Q}$ we get that
$$\mathbb{Q}[t]/(t^n - 1) \cong \prod_{d | n} \mathbb{Q}[t]/\Phi_d(t) \cong \prod_{d | n} \mathbb{Q}(\zeta_d)$$
is a product of cyclotomic fields, one for each divisor of $d$. 
A: Denote the elementwise multiplication operation by $\cdot$ and the one induced by the identification with $A[t] / (t^2 - 1)$ by $\ast$. These rings have respective identity elements $1_{\cdot} = (1, 1)$ and $1_{\ast} = (1, 0)$.

First, consider the special case in which $A$ is a field. If there is a ring isomorphism, it must map zero divisors to zero divisors, and some easy algebra shows that


*

*the zero divisors with respect to $\cdot$ are the elements of the form $(a, 0)$ or $(0, a)$, $a \in A$, and

*the zero divisors with respect to $\cdot$ are the elements of the form $(a, a)$ and $(a, -a)$, $a \in A$. (Indeed, we have $(a, a) \ast (a, -a) = (0, 0).)


This suggests some natural candidates for isomorphisms $(A \times A, +, \cdot) \to (A \times A, +, \ast)$, namely the linear maps $\phi$ characterized by $$(a, 0) \mapsto (\lambda a, \lambda a), \qquad (0, \lambda a) \mapsto (-\lambda a, \lambda a)$$ for $\lambda \in A$---that is, the maps
$$\phi: (a, b) \mapsto (\lambda(a + b), \lambda(-a + b))$$---and some variants thereof given by introducing appropriate sign changes. Now, if $\phi$ is an isomorphism, we must have that $\phi(1_{\cdot}) = \phi((1, 1)) = (2\lambda, 0)$ coincides with $1_{\ast} = (1, 0)$. I do not know whether the two operations define isomorphic rings for any field of characteristic $2$ (or ring of characteristic $2$ for that matter).
If $\operatorname{char} A \neq 2$, then for $\phi$ to be an isomorphism we must take $\lambda = 2^{-1}$. We can then verify directly that $\phi((a, b) \cdot (c, d)) = \phi(a, b) \ast \phi(c, d)$, that is, that $\phi$ is indeed an isomorphism.
On the other hand, if $\operatorname{char} A = 2$, then $\phi$ is not an isomorphism, and indeed, the two rings need not be isomorphic. One can easily see that this already occurs for $A = \Bbb F_2$: There are six ordered pairs $((a, b), (c, d)) \in (\Bbb F_2 \times \Bbb F_2) \times (\Bbb F_2 \times \Bbb F_2)$ whose product under $\cdot$ is the additive identity $(0, 0)$, but there are only five ordered pairs whose product under $\ast$ is $(0, 0)$.
Our construction only uses that $A$ is a field when setting $\lambda = 2^{-1}$ (a field axiom guarantees that all nonzero elements have multiplicative inverses). So, for a general nontrivial ring $A$ in which $2$ is invertible, the map $\phi$ given by the above formula is again a ring homomorphism.
The simplest nontrivial ring of characteristic $\neq 2$ in which $2$ is not invertible is $\Bbb Z_4$, and one can see, again by counting the number of ordered pairs whose product is the additive identity, that the two product structures on $\Bbb Z_4 \times \Bbb Z_4$ are nonisomorphic, and so $\operatorname{char} A \neq 2$ is not alone a sufficient condition to guarantee isomorphism of the two ring structures on $A \times A$ for a nontrivial ring $A$.
