Prove if $I$ is an ideal of $R$ ring, then $(I \cup \{x\})/(x) \cong I$.

Question

I am trying to prove the statement above where $(x)$ is the principal ideal generated by the polynom $x$ in $R[x]$, and similarly with $(I \cup \{x\})$. Additionally, I have to prove that $R[x]/(I \cup \{x\}) \cong R/I$ with the help of this statement. For this part, I know that $R[x]/(x) \cong R$ and I should use the 3rd (or maybe 2nd, I don't know which one it is in English) isomorphism theorem, but I'm clueless with the initial statement. I appreciate any sort of help.

Answer 1

The set $(I \cup \{x\})$ is a set of polynomials over $R$, so define a function $\varphi$ from $(I \cup \{x\})$ to $I$ which simply selects the constant term of this polynomial.

This is a homomorphism (this is easy to prove), surjective and its kernel is $(x)$, which proves your proposition through the first isomorphism theorem (sometimes called the homomorphism theorem)

Answer 2

Hint: Use the first isomorphism theorem for the natural map $R[x]\to R/I$ sending $x\mapsto 0$.