Let $B$ be an $A$-algebra. (That is, $\varphi:A\to B$). Suppose $y\in B$ zeroes a normalized polynomial $0\ne f\in A[x]$. (That is, $y$ is integral in $A$). Show that $L:=\varphi(A)[y]$ is finite above $A$. ($L$ is closure of $\{\varphi(a):a\in A\}\cup\{y\}$ under $+,-,\cdot$).
In the solution:
Let $$0\ne x^n+c_{n-1}x^{n-1} +\dots+c_0\in A[x]$$ which zeroed in $y$. Then $$ 0=y^n+\varphi(c_{n-1})y^{n-1}+\dots+\varphi(c_0) $$ But I don't understand the last equlity. I guess that the meaning is that $$ 0=\varphi (0)=\varphi(f(y)) \\=\varphi (y^n+c_{n-1}y^{n-1}+\dots+c_0) \\=\varphi(y)^n+\varphi(c_{n-1})\varphi(y^{n-1})+\dots+\varphi(c_0) $$ But the last transition is not valid since $\varphi(y)$ is not defined! ($y\notin A =\text{Domain}(\varphi)$).
I would like for an explanation, thanks in adence.
To evaluate a polynomial $f\in A[x]$ at a point $y\in B$, one uses the universal property of the polynomial ring: there is a unique homomorphism $\bar\varphi_y: A[x] \to B$ with $\bar\varphi_y|_A=\varphi$ and $\bar\varphi_y(x)=y$. Then, one usually writes $f(y)=\bar\varphi(f)$. Consequently, $$ 0 = f(y) = \bar\varphi_y(f) = y^n + \varphi(c_{n-1})y^{n-1}+\cdots+\varphi(c_0). $$