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
$$
R/\bigl\langle a,b \bigr\rangle = R/\left(Ra+Rb\right).
$$

 

[elias001]

@fresh_42 in my original post, in screenshot 2, where it says

$\bar{b}=\frac{(a,b)}{(a)}.\\\\$



In the context of ##\frac{\Bbb{Z}[x]}{\langle X^2+1, X-2\rangle}
\cong\frac{\mathbb{Z}}{\langle i-2\rangle}\cong \mathbb{Z}_5,\\\\$Can we show$\bar{X-2}=\frac{\langle X^2+1, X-2\rangle}{\langle X^2+1\rangle}.\\\\$Also for the fourth isomorphism theorem, in the context of$
\frac{\Bbb{Z}[x]}{\langle X^2+1, X-2\rangle}
\cong\frac{\mathbb{Z}}{\langle i-2\rangle}\cong
\mathbb{Z}_5,\\\\## can we see what the forward and inverse image map look like in order to show the fourth isomorphism theorem?

 

[fresh_42]

I still have problems with that notation. What does that horizontal line even mean? And the italic tags in your posts damage the LaTeX code, which makes it very difficult to read.

 

[elias001]

@fresh_42 you know in the third isomorphism $(R/I)/(I/J)\cong R/J$, $R=\Bbb{Z}[X], I=\langle X^2+1, X-2\rangle, J=\langle X^2+1\rangle, b=\langle X-2\rangle.$ According to example 4.1 in screenshot 2,, then $\bar{\langle b\rangle}=\bar{X-2}= \langle X^2+1, X-2\rangle/\langle X^2+1\rangle.$ I am not sure how to remove the italics in my previous reply.