- #1
Joelly
- 3
- 0
In summary, the function that maps $a+b\sqrt{2}$ to $\begin{bmatrix}a & 2b \\ b & a \end{bmatrix}$ is an obvious "guess" and can be proven to be an isomorphism by showing that it is "one to one" and "onto" and that it "preserves the operation" for multiplication and addition. This is done by comparing the products and sums of two members in the first set and their corresponding mapped matrices.
- #2
HOI
- 921
- 2
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.
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.
- #3
Joelly
- 3
- 0
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