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Is any extension ring $S \supset R$ an $R$-algebra?

In our lecture note an $R$-algebra $A$ is defined as follows $:$

An $R$-algebra $A$ is a ring $A$, which is also an $R$-module satisfying the condition

$$a(xy)=(ax)y=x(ay),\ a \in R,\ x,y \in A.$$

It is clear that $S$ is a ring along with an $R$-module with the usual addition and scalar multiplication. But how does the scalar multiplication become bilinear unless $R \subset Z(S)$ where $Z(S)$ denotes the center of $S$.

I think the assertion made in our lecture note is false. Please check it.

Thank you in advance.

Arnab Chattopadhyay.
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    Note that setting $y=1$ you do get that $ax=xa$ for $x\in A$ and $a\in R$, so that $R$ must have image in $Z(S)$. – Pedro Apr 28 '18 at 16:03
  • The compatibiliy relation for algebra multiplication certainly fails to hold for the case $S=M_n(\mathbb C)$ and $R$ is the diagonal matrices. – Aweygan Apr 28 '18 at 16:04
  • Would you please be more explicit @Pedro Tamaroff? How do I assume the bilinearity property without establishing it (as you said setting y=1...)? Also how do we assume $1 \in A$? – Arnab Chattopadhyay. Apr 28 '18 at 16:11

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No: take for example $\mathbb C \subseteq \mathbb H$, the complex numbers in the quaternions. (This is an example if what Pedro was getting at in the comments.)

The axioms of an $R$ algebra $A$ require that $R$ is contained in the center of $A$, which is not the case for this example.

rschwieb
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