Is this version of evaluation homomorphism theorem correct?

Question

The theorem comes from Kostrikin's Introduction to Algebra and goes as follows:

If $A$ (a commutative ring with an identity) is a subring of a commutative ring $R$, then for every element $t \in R$ there exists exactly one homomorphism $\Pi_t: A[X] \rightarrow R$, such that $\Pi_t(X)=t$ and $\Pi_t(a)=a$ for every $a \in A$.

I'm having some doubts about my understanding of this. If we set $A=\{(a, 0): a \in \mathbb{Z}\}$, $R=\mathbb{Z}^2$ and $t=(0,1)$, then these fulfil all requirements of the theorem if we define operations on them as $(a, b) + (c, d) = (a+c, b+d)$ and $(a, b)(c,d)=(ac, bd)$. Therefore the described homomorphism should exist. However, my "counterexample" to that would be:

$(0,0)=(1,0)(0,1)=\Pi_{(0,1)}((1,0))\Pi_{(0,1)}(X)=\Pi_{(0,1)}((1,0)X)=\Pi_{(0,1)}(X)=(0,1)$

So that shows that something here isn't quite right.

Answer 1

The problem step seems to be $\Pi_{(0,1)}((1,0)X)=\Pi_{(0,1)}(X)$. Indeed, $(1,0)$ is not the multiplicative identity in $\Bbb Z^2$; that role belongs to $(1,1)$.

Answer 2

The issue is when you write

$$\Pi_{(0,1)}((1,0)X)=\Pi_{(0,1)}(X)$$

I imagine that you suppose that $(1,0)X=X$. Which might not be the case.