The following is taken from Abstract Algebra A Comprehensive Introduction by: Lawerence and Zorzitto.
Background
The ideal in a commutative ring $R$ generated by eleements $a_1,\dots, a_n$ is denoted by the wedge bracket notation:
$$\langle a_1,\dots, a_n \rangle$$
and sometiems the more informative noatation
$$Ra_1+Ra_2\dots +Ra_n.$$
Proposition 4.32 (Euclidean division with polynomials). If $R$ is a commutative ring and $g$ is a non-zero polynomial in $R[X]$ such that its leading coefficient is a unit of $R$, then for eveyr $f$ in $R[X]$ there exist unique polynomials $q,r$ in $R[X]$ so that
$$f=gq+r\quad\text{and}\quad \text{deg }r<\text{deg }g\text{ or }r=0.$$
In particular, if $R$ is a field, then the above holds for all non-zero $g$.
Propositiion 4.49 (Correspondence theorem).: Let $\phi: R\to A$ be a surjective ring homomorphism. (Here $R,A$ need not be commutative.) Let $\mathcal{J}$ be the family of ideals of $R$ containing $\text{ker }\phi$. and $\mathcal{L}$ the family of ideals of $A$.
(1) If $J \in\mathcal{J}$, then $\phi(J) \in\mathcal{L}$
(2) If $L \in\mathcal{L}$, then $\phi^{-1}(L) \in \mathcal{J}$
(3) If $J \in\mathcal{J}$, then $\phi^{-1}(\phi(J))=J.$
(4) If $ L\in\mathcal{L}$, then $\phi(\phi^{-1}(L))=L.$
(5) The mapping $\mathcal{J}\to \mathcal{L}$ defined by $J\mapsto \phi(J)$ is a bijection whose inverse is the mapping $\mathcal{L}\to \mathcal{J}$ defined by $L\mapsto \phi^{-1}(L)$
(6) If $J\in \mathcal{J}$, then $R/J\cong A/\phi(J) $
An Illustration of the Correspondence Theorem
The isomorphism in item 6 of the correspondence theorem can be useful in detectin the nature of some quotient rings. Here is a scenerario which could occur.
$\quad$ Say $R$ is a commutative ring and $a,b\in R$. What can Proposition 4.49 reveal about the quotient ring $R/\langle a,b\rangle$? Well, the canonical projection $\phi: R\to R/\langle b\rangle$ is a surjective homomorphism, and its kernel is the principal ideal $\langle b\rangle$. The ideal $\langle a,b \rangle$ contains $\langle b\rangle$. By a simple verification the image $\phi(\langle a,b \rangle)$ is the ideal $\langle \phi(a),\phi(b)\rangle$ inside $R/\langle b\rangle$. Since $\phi(b)=0$ the image $\phi(\langle a,b\rangle)$ is the principal ideal $\langle \phi(a)\rangle$. According to the correspondence theorem
$$R/\langle a,b\rangle\cong (R/\langle b\rangle)/\langle a\rangle).$$
The right side is a succession of quotients of two principal ideals, which more readily be deciphered.
$\quad$ For example, let us describe the quotient ring $\Bbb{Z}[X]/\langle X^2+1,X-2 \rangle$. Start with the substitution map $\phi:\Bbb{Z}[X]\to \Bbb{Z}$ given by $f\mapsto f(2).$ This is a surjection and its kernel is the ideal $\langle X-2\rangle$. Indeed, $\langle X-2\rangle\subset \text{ker }\phi$ because $X-2$ has a root at $2$. On the other hand $f\in \text{ker }\phi$. Since $X-2$ is monic, Proposition 4.32 gives a polynomials $q,r$ in $\Bbb{Z}[X]$ such that $f=(X-2)q+r$ and $r$ is constant. From $f(2)=0$ it follows that $r=0,$ and thus $f\in \langle X-2\rangle$. $\quad$ The correspondence theorem gives a bijection between the ideals of $\Bbb{Z}[X]$ containing $\langle X-2\rangle$ and the ideals of $\Bbb{Z}$. Now $\langle X^2 +1, X-2\rangle$ contains $\langle X-2\rangle$. By item 6 of the correspondence theorem
$$\Bbb{Z}[X]/\langle X^2 +1, X-2\rangle\cong \Bbb{Z}/\phi(\langle X^2 +1, X-2\rangle).$$
In $\Bbb{Z}$ the ideal
$$\phi(\langle X^2 +1, X-2\rangle)=\langle \phi(X^2+1),\phi(X-2)\rangle=\langle 2^2+1,2-2\rangle=\langle 5\rangle$$
Thus
$$\Bbb{Z}[X]/\langle X^2 +1, X-2\rangle\cong \Bbb{Z}/\langle 5\rangle=\Bbb{Z}_5.$$
The original quotient is isomorphic to the finite field $\Bbb{Z}_5$.
$\quad$ Let us see what happens if we carry out the process in a different order. Take the substitution map $\psi:\Bbb{Z}[X]\to\Bbb{Z}[i]$ given by $f\mapsto f(i)$. This is a surjective homomorphism and its kernel is $\langle X^2+1 \rangle$. Indeed, $\langle X^2+1 \rangle\in \text{ker }\psi$ because $i$ is a root of $X^2+1$. On the other hand let $f\in \text{ker }\psi$. Since $X^2+1$ is monic, Euclidean division gives polynomials $q,r$ in $\Bbb{Z}[X]$ such that $f=(X^2+1)q+r$ and $r$ is of the form $a+bX$ for some $a,b$ in $\Bbb{Z}$. Since $f(2)=0$ we get $r(2)=a+bi=0.$ By the nature of complex numbers $a=b=0$, and thus $r=0$. Having seen that $f=(X^2+1)q$ it follows that $\text{ker }\psi=\langle X^2+1 \rangle$. By item 6 of the correspondence theorem
$$\Bbb{Z}[X]/\langle X^2 +1, X-2\rangle\cong \Bbb{Z}[i]/\psi(\langle X^2 +1, X-2\rangle) =\Bbb{Z}[i]/\langle \psi(X^2+1),\psi(X-2)\rangle=\Bbb{Z}[i]/\langle i-2\rangle.$$
Putting all of this together we come up with the less than obvious discovery that
$$\Bbb{Z}[i]/\langle i-2\rangle\cong \Bbb{Z}_5.$$
Questions
I have some questisons about the above exposition on the illusration of the correspondence theorem. I often see post such as: Describe the ring structure quotient ring $Z[x,y]/(x^2,y^2,2)$. and isomorphism $\mathbb{C} [X,Y]/(X^2 + Y^2 - 1)$. I often wonder there is some slight missing justifications especially in the first reference post in the sense if we are given a quotient ring with its ideal that is finitely generated, then there such problem. Anyways, onto my questions (a) and (b) below:
(a) By a simple verification the image $\phi(\langle a,b \rangle)$ is the ideal $\langle \phi(a),\phi(b)\rangle$ inside $R/\langle b\rangle$. Since $\phi(b)=0$ the image $\phi(\langle a,b\rangle)$ is the principal ideal $\langle \phi(a)\rangle$. According to the correspondence theorem
$$R/\langle a,b\rangle\cong (R/\langle b\rangle)/\langle a\rangle).$$
and
(b) $\phi(\langle X^2 +1, X-2\rangle)=\langle \phi(X^2+1),\phi(X-2)\rangle=\langle 2^2+1,2-2\rangle=\langle 5\rangle$
For (a) I understand that "$\phi(\langle a,b \rangle)$ is the ideal $\langle \phi(a),\phi(b)\rangle$" is due to Proposition 4.49 (1) since $\langle a,b \rangle\in \mathcal{J}$ and $\langle \phi(a),\phi(b)\rangle\in \mathcal{L}$. But which is inside $R/\langle b\rangle$? $\phi(\langle a,b \rangle)$ or $\langle \phi(a),\phi(b)\rangle$? And how?
When it says "Since $\phi(b)=0$ the image $\phi(\langle a,b\rangle)$ is the principal ideal $\langle \phi(a)\rangle$", I am guessing "$\phi(b)=0$" is part of the or is the kernel of some evalulation map. And $\langle \phi(a)\rangle=\{\langle\phi(\langle a,b\rangle)\rangle\mid \langle a,b\rangle\in \mathcal{J}\}$, but how do $\phi(b)=0$ relate to the principal ideal $\langle \phi(a)\rangle$? Lastly how does the correspondence thorem give rise to the following isomorphism: $R/\langle a,b\rangle\cong (R/\langle b\rangle)/\langle a\rangle)$? I understand that there should be some map going from $R/\langle b\rangle\to R/\langle a,b\rangle$ with $\langle a\rangle$ being the kernel of the said map? But how do: By a simple verification the image $\phi(\langle a,b \rangle)$ is the ideal $\langle \phi(a),\phi(b)\rangle$ inside $R/\langle b\rangle$. Since $\phi(b)=0$ the image $\phi(\langle a,b\rangle)$ is the principal ideal $\langle \phi(a)\rangle$. along with the correspondence theorem give rise to the isomorphism: $R/\langle a,b\rangle\cong (R/\langle b\rangle)/\langle a\rangle)$? I think the map should be $r+\langle b\rangle\mapsto r+\langle a,b\rangle$.
For (b) what justifies the step $\langle 2^2+1,2-2\rangle=\langle 5\rangle$? I mean which paprt of the correspondence theorem allows one to simply plug in the numbers $2$ into $\langle X^2+1,X-2\rangle$.
Thank you in advance
