I'm struggling to understand the definition of ideals in ring homomorphisms generated by a set.
If $R$ is commutative and has a $1$, then Ideal of $R$ generated by a subset $A$ of $R$:
$$⟨ A ⟩ = \{r_1a_1+\dotsb+r_na_n\mid r_i\in R, a_i\in A, n\in \mathbb{N}\}.$$
Now if $R$ has a $1$ isn't it sufficient to always use $⟨1⟩$ to express each element in the ideal?
$$⟨ 1 ⟩ = \{1r \mid r \in R\} = R$$
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$\begingroup$I'm not sure what you're asking. If $1$ is in your ideal $I$ then yes, as you wrote, $I = R$, so it's not a proper ideal.
Maybe an example of an ideal generated by a set helps: Let $R = \mathbb Z$ and $A = \{7\}$. Then $\langle A \rangle = 7 \mathbb Z$.
If $R = K[x,y]$ for some field $K$ and $A = \{x,y\}$ then $I = \langle x,y \rangle $ is the set of all polynomials with no constant term. On the other hand, if $A = \{2x \}$ then $\langle 2x \rangle$ is the set of all polynomials with no constant term and with only even coefficients.
Hope this helps. If I misunderstood your question just drop me a comment.
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