Contraction of quotient ideal is quotient of contractions?

Question

Let $\mathfrak a,\mathfrak b$ be ideals in a ring $A.$ The quotient of $\mathfrak a$ and $\mathfrak b$ is $(\mathfrak a:\mathfrak b)=\{x\in A:x\mathfrak b\subseteq \mathfrak a\}$ and if $f:B\to A$ is a ring homomorphism, then the contraction of $\mathfrak b$ is $\mathfrak b^c=f^{-1}(\mathfrak b).$

In one of the exercises in the first chapter of Atiyah-Macdonald (commutative algebra), one is asked to show that $(\mathfrak a:\mathfrak b)^c\subseteq(\mathfrak a^c:\mathfrak b^c).$ I did this perfectly fine, but one of the remarks after the exercises implies to me that this should be an equality, which doesn't seem right.

Explicitly, they remark "Thus extension is closed under the operations of ideal addition and multiplication, while contraction is closed under [intersections, radicals, and quotients]." Indeed, part of the exercise was to show that the extension of the sum is the sum of the extensions, etc. This is the only one (of those mentioned in the remark) in which equality was not shown.

Is the remark a mistake, or is there a typo in the exercise and I'm wrong?

Answer 1

It seems this is wrong. In this example, page 8, a counter example for the inclusion $(\mathfrak a:\mathfrak b)^c\overset{?}{\supseteq}(\mathfrak a^c:\mathfrak b^c)$ is provided. Choose $f\colon\mathbf Z\hookrightarrow\mathbf Z[i]$ with ideals $\mathfrak a=(2+i)$, $\mathfrak b=(2-i)\subseteq\mathbf Z[i]$; note that these are prime. Then $\mathfrak a^c=\mathfrak a\cap\mathbf Z=(5)=\mathfrak b^c$, so $(\mathfrak a^c:\mathfrak b^c)=\mathbf Z$. On the other hand, $(\mathfrak a:\mathfrak b)=(\mathfrak a)$ because both are prime ideals, so $(\mathfrak a:\mathfrak b)^c=(5)\subsetneq (\mathfrak a^c:\mathfrak b^c)$.

Answer 2

To flesh out the helpful comment by user26857, everything in the book is correct, it just requires a little work to see. The claim is that $(\mathfrak a^c:\mathfrak b^c)$ is a contraction, namely $(\mathfrak a:\mathfrak b^{ce})^c.$ This is rather easy to see, I guess. \begin{align*} (\mathfrak a^c:\mathfrak b^c) &= \{x\in B : xf^{-1}(\mathfrak b)\subseteq f^{-1}(\mathfrak a)\}\\ &=\{x\in B : xy\in f^{-1}(\mathfrak a) \text{ for all }y\in f^{-1}(\mathfrak b)\}\\ &=\{x\in B : f(xy)=f(x)f(y)\in \mathfrak a \text{ for all }y \text{ with } f(y)\in\mathfrak b\}\\ &=\{x\in B : f(x)z\in \mathfrak a \text{ for all }z\in f(f^{-1}(\mathfrak b))\}\\ &=\{x\in B : f(x)z\in \mathfrak a \text{ for all }z\in (f(f^{-1}(\mathfrak b)))\}\\ &=\{x\in B : f(x)z\in \mathfrak a \text{ for all }z\in \mathfrak b^{ce}\}\\ &=\{x\in B : f(x)\in (\mathfrak a:\mathfrak b^{ce})\}\\ &=(\mathfrak a:\mathfrak b^{ce})^c \end{align*} I guess is this a lesson on just how terse this book actually is.