Let $A$ be a ring and $A/I$ a quotient ring where $I$ is a ideal of $A$.
My teacher said that if there's an injective homomorphism $\varphi:A\to A/I$, $x\mapsto \overline{x}$ (an inclusion map), then I can consider $x$ and $\varphi(x)=\overline{x} $ as equal elements. That is, if I'm working with the equation $y+x=z$, with $y,z\in A$, then I can, if it's convenient, swap $x$ by $\overline{x}$ and get $y+\overline{x}=z$.
I would like to confirm if this is really true since it is strange to consider an element $x$ equal to its equivalence class $\overline{x}$, for example.
I wonder if this is the same thing as considering $1=\sqrt{1+\frac{\sqrt{3}}{2}}-\sqrt{1-\frac{\sqrt{3}}{2}}$ although $1$ and $\sqrt{1+\frac{\sqrt{3}}{2}}-\sqrt{1-\frac{\sqrt{3}}{2}}$ are quite different expressions.
My question is: can I really, if I get an inclusive map $\sigma:A\to B,\, x\mapsto \sigma(x)$, consider $x$ equals to $\sigma(x)$?
EDIT: Let $K$ be a field. Change the map $\varphi:A\to A/I,\, x\mapsto \overline{x}$ by the map $\varphi' :K\to K[X],\, a\mapsto a(X)$ in which $a(X)=a$ is the constant polynomial of $K[X]$ equal to $a$. Note that $\varphi'$ is a injective homomorfism.
If you have an inclusion map $A \to B$, you can consider $A$ to be a substructure of $B$. In that sense, yes, you can consider elements of $A$ to be elements of $B$. This is true for any algebraic structure, not only rings.
Be careful though: the names you used for elements of $A$ a priori will not necessarily be the same names they have in $B$. For a simple example, there is a group homomorphism $\mathbb{Z} \to \mathbb{Z}$ given by multiplication by $2$. It's injective, so it should follow your rule. But it makes sense to view it as inclusion only if we name the elements of the domain $\{0,2,4,\dots\}$ instead of $\{0,1,2,\dots\}$.