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An obvious "guess", practically forced by the way the problem is stated, is the function that maps $a+ b\sqrt{2}$ to $\begin{bmatrix} a & 2b \\ b & a \end{bmatrix}$. Is that an isomorphism? That is, is it "one to one", is it "onto", and does it "preserve the operation".

The first two are simple. To prove that this "preserves the operation" you need to take two members of the first set, $a+ b\sqrt{2}$ and $x+ y\sqrt{2}$, which map to $\begin{bmatrix}a & 2b \\ b & a\end{bmatrix}$ and $\begin{bmatrix}x & 2y \\ y & a\end{bmatrix}$ and show that the product $(a+ b\sqrt{2})(x+ y\sqrt{2})$ maps to the product $\begin{bmatrix}a & 2b \\ b & a \end{bmatrix}\begin{bmatrix}x & 2y \\ y & x\end{bmatrix}$.

Do those two multiplications and compare them.

To determine whether they are also isomorphic under addition, add those pairs and compare them.
 
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I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
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