How do we know that the canonical surjection $\mathbb{Z}[\sqrt{2}] \rightarrow \mathbb{Z}[\sqrt{2}]$ is injective?

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Let $R$ denote a commutative ring. Consider a polynomial $P(x) \in R[x]$ and a commutative $R$-algebra $S$. Suppose we're given $a \in R$ such that $P(a) = 0$. There's an $R$-algebra homomorphism $R[x]/P(x) \rightarrow S$ given by mapping $x$ to $a$. My question is:

General Question. What assumptions do we need (e.g. on $S$) for this to be injective?

Here's an example of why we might care:

Let $\mathbb{Z}[\sqrt{2}]$ denote the quotient of $\mathbb{Z}[x]$ by the polynomial $x^2-2$.

Let $\mathbb{Z}[\sqrt{2}]$ denote the smallest subring of $\mathbb{R}$ containing the unique $a \in \mathbb{R}$ satisfying the following conditions: $$a^2-2 = 0, \qquad a>0.$$

There's a surjection $\mathbb{Z}[\sqrt{2}] \rightarrow \mathbb{Z}[\sqrt{2}]$ given by mapping $x$ to $a$.

Specific Question. How can we show that this map injective?

To be clear, the actual question I'd like answered is the general one, not the specific one.

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Simply show that the quotient of the polynomial ring $\mathbb Z[x]$ by $x^2-2$ has the classes of $1$ and of $x$ as a basis as an abelian group — this follows immediately from the fact that the polynomial is monic. Then just check by hand what you want.