Does the cyclic algebra construction preserve surjective maps and direct products?

Let $R$ be a commutative ring with automorphism $\theta$ such that $\theta^n=Id$. We define the cyclic algebra $\mathbf{C}_n(R,\theta)$ to be the two-sided free $R$-module of rank $n$ with basis $\{1,x,\ldots,x^{n-1}\}$

$\mathbf{C}_n(R,\theta)=R\oplus R\cdot x\oplus\cdots\oplus R\cdot x^{n-1}$

such that $y^n=1$ and $x^i r=\theta^i(r)x^i$ for $0\leq i\leq n-1$. I need to decide whether the following two statements are correct:

1) Suppose $S$ is another commutative ring with automorphism $\tau$ and such that $\varphi:R\to S$ is a surjective ring homomorphism where $\varphi\circ\theta=\tau\circ\varphi$. Is the induced map $\tilde{\varphi}:\mathbf{C}_n(R,\theta)\to\mathbf{C}_n(S,\tau)$ surjective?

2) With $S$ as above, is it true that $\mathbf{C}_n(R\times S,\theta\times\tau)\cong\mathbf{C}_n(R,\theta)\times\mathbf{C}_n(S,\tau)$?

The first one seems obviously true since we can represent elements of $\mathbf{C}_n(S,\tau)$ by $(s_1,\ldots,s_n)$ and each $s_i$ can be written as $\varphi(r_i)$.

However the second point seems to be eluding me and I am unsure whether I am overthinking it. It seems to me that it must be true, but I can't seem to show this.