I Determining isomorphism for ##\frac{R}{(a, b)}##

[elias001]

TL;DR Summary

Which isomorphism theorems should be used when determining what ring is isomorphic to $\frac{R}{(a, b)}$.

The screenshots below are taken from the two books Algebra: Chapter 0 by: Paolo Aluffi and Abstract Algebra A comprehensive Introduction by Lawrence and Zorzitto. The first two screenshots are from Aluffi's text, while the last four are from Lawrence and Zorzitto.

Screenshot 1

Screenshot 2

Screenshot 3

Screenshot 4

Screenshot 5

Screenshot 6

Given a homomorphism $\varphi:R\to A$, in the case of ideals in commutative rings, if I want to determine what the quotient ring $\frac{R}{\langle a, b\rangle}$ is isomorphic to, in the first two screenshot from the Aluffi's text, he states that reader can make use of Proposition 3.11, the third isomorphism theorem. His justification is that: $frac{\frac{R}{(a)}}{(\bar{b})}\cong\frac{R}{(a,b)}\quad (\bar{b}=\frac{(a,b)}{(a)}.$

However, in the next four screenshots, from Lawrence and Zorzitto's text, they suggest to the reader that the fourth isomorphism or the Correspondence theorem can be used due to the following reasons: $\frac{R}{\langle a, b\rangle}\cong \frac{\frac{R}{\langle b\rangle}}{\langle \phi(a)\rangle}$. The example they gave is that of $\frac{\Bbb{Z}[x]}{\langle X^2+1, X-2\rangle}$.

Am I to conclude that say if I need to cite a reasons for resolving which other ring a quotient ring$\frac{R}{\langle a, b\rangle}$ is isomorphic to, I can cite either the correspondence theorem or the third isomorphism theorem?

Thank you in advance

 

[fresh_42]

The difficulty is - as usual - to make sure which ideal lives in which ring. We do not know wether $\bigl\langle a \bigr\rangle \subseteq \bigl\langle b \bigr\rangle$ or $\bigl\langle b \bigr\rangle \subseteq \bigl\langle a \bigr\rangle$ so we cannot simply factor them. And in the example, neither is true. This forces us to establish such a conclusion before we can build the quotient.

Let me resolve the vague notation by writing $\bigl\langle a,b \bigr\rangle =Ra+Rb.$ Then
$$
R/\bigl\langle a,b \bigr\rangle =R/\left(Ra + Rb\right)\cong \left(R/Rb \right)/\left(\left(Ra+Rb\right)/Rb\right)=R/\bigl\langle b \bigr\rangle \, / \, \left(\bigl\langle a,b \bigr\rangle / \bigl\langle b \bigr\rangle\right).
$$
I know this as the second isomorphism theorem, and the English Wikipedia refers to it as Theorem C (point 5). If you phrase it with ring homomorphisms, then you get a mixture of several isomorphism theorems (theorems A and B on the English Wikipedia page).

More important than knowing which is called what is understanding the mechanisms. You can only build a quotient if you have an ideal (apart from special cases that do not play a role here). E.g. $\bigl\langle X^2+1 \bigr\rangle$ isn't an ideal in $\bigl\langle X-2 \bigr\rangle$ nor the other way around.

The example asks for
$$
\mathbb{Z}[x]/\bigl\langle x^2+1,x-2 \bigr\rangle.
$$
The ideal notation means
$$
\bigl\langle x^2+1,x-2 \bigr\rangle=\mathbb{Z}[x]\cdot(x^2+1)+\mathbb{Z}[x]\cdot(x-2).
$$
Let me abbreviate the ideal and rings by setting
$$
R=\mathbb{Z}[x]\, , \,I=\mathbb{Z}[x]\cdot (x^2+1)\, , \,J=\mathbb{Z}[x]\cdot (x-2)
$$
By the (second) isomorphism theorem, e.g. 3.11, we get
\begin{align*}
R/\left(I+J\right)&\cong \left(R/J\right)/\left(\left(I+J\right)/J\right).
\end{align*}
The ring $R/J$ consists of all integer polynomials where we have identified $x=2,$ leaving us with an isomorphic copy of $\mathbb{Z}.$ Now, what ring is $\left(I+J\right)/J?$ We know that $J/J=\{0\},$ but $J\not\subseteq I,$ so we cannot build $I/J.$ Instead, we have to take all polynomials in
$$
I+J=\mathbb{Z}[x]\cdot(x^2+1)+\mathbb{Z}[x]\cdot(x-2)
$$
modulo $x-2.$ This means we replace all occurrences of $x$ by $2$ and get polynomials from $\mathbb{Z}[2]\cdot 5=5\cdot\mathbb{Z}.$ Thus,
$$
R/\left(I+J\right)\cong \mathbb{Z}/5\mathbb{Z}\cong \mathbb{Z}_5.
$$
The second process simply exchanges the roles (definitions) of $I$ and $J.$ Here, we have
\begin{align*}
R/\left(I+J\right)&\cong \left(R/I\right)/\left(\left(I+J\right)/I\right).
\end{align*}
The ring $R/I$ is isomorphic to $\mathbb{Z}[\mathrm{\,i\,}].$ So what is $\left(I+J\right)/I?$ We still have all polynomials
$$
I+J=\mathbb{Z}[x]\cdot(x^2+1)+\mathbb{Z}[x]\cdot(x-2)
$$
which we now take modulo $x^2+1.$ The first term vanishes, and the integer coefficients of the second term become $\mathbb{Z}[\,\mathrm{i}\,]$ plus that we then have to consider the ideal $I+J$ modulo $x^2+1,$ i.e., we set $x=i,$ turning $\left(I+J\right)/I$ into an isomorphic copy of $\mathbb{Z}[\,\mathrm{i}\,]\cdot (i-2)$ and finally
$$
R/\left(I+J\right)\cong \mathbb{Z}[\,\mathrm{i}\,]/ \mathbb{Z}[\,\mathrm{i}\,]\cdot (i-2).
$$

Both statements together result in
$$
R/\left(I+J\right)\cong \mathbb{Z}[x]/\bigl\langle x^2+1,x-2 \bigr\rangle \cong \mathbb{Z}/5\mathbb{Z}\cong \mathbb{Z}_5=\mathbb{Z}/\bigl\langle 5 \bigr\rangle \cong \mathbb{Z}[\,\mathrm{i}\,]/ \mathbb{Z}[\,\mathrm{i}\,]\cdot (i-2)=\mathbb{Z}[\,\mathrm{i}\,]/\bigl\langle i-2 \bigr\rangle .
$$

Note that the brackets always mean something different, depending of where the ideals live in. It is therefore better to write, for example
$$
R/\bigl\langle a,b \bigr\rangle = R/\left(Ra+Rb\right).
$$