I am solving the following exercise and will appreciate a little help with some steps. This is the exercise and my solution process is below.
1. Show that the reducts are isomorphic
To show an isomorphism of algebras, we need to show there is a homomorphism $H_1: A \to A$ with an inverse homomorphism $H_2: A \to A$, $H_2 := (H_1)^{-1}$. $H_1$ and $H_2$ must be bijective.
The $H_1$ and $H_2$ must preserve the operations of the reducts. Let´s start with $H_1$.
Define $H_1$ as:
\begin{array}{ll}
(0,0) & x = (0,0) \\
b & x = a \\
a & x = b \\
Id & x \in K – a, b, (0,0) \\
Id & x \in K´\\
\end{array}
$H_1(x *_1 y) = H_1(x) *_2 H_1(y)$ must be true for any $x$ and $y$ from $A$.
To see whether this holds, let us look at the two possible cases.
x and y are in K´
Then the operation $*_1$ gives $a$ everytime and $*_2$ gives $b$. So we got to check that
$H_1(a) = H_1(x) *_2 H_1(y)$. According to the definition,
LS = $H_1(a) = b$
RS = $H_1(x) *_2 H_1(y) = x *_2 y = b$. Hence $LS = PS$.
x and y are not in K´
Then $LS = H_1((0,0)) = (0,0)$.
$PS = H_1((0,0)) *_2 H_1((0,0)) = (0,0) *_2 (0,0) = (0,0).$
Now check the $H_2$ (inversion).
x and y are in K´
$LS = H_2(x *_2 y) = H_2(b) = (H_1)^{-1}(b) = a$
$PS = H_2(x) *_1 H_2(y) = x *_1 y = a$
x and y are not in K´
$LS = H_2(x *_2 y) = H_2((0,0)) = (0,0)$
$PS = H_2(x) *_1 H_2(y) = (0,0) *_1 (0,0) = (0,0)$.
Hence, we have bijective homomorphism with bijective inversion, so there exists the desired isomorphism.
My questions:
A) How do I know that my definition of $H_1$ (and its inverse) is correct?
B) How to proceed with the second part about ring isomorphisms?
If this would be correctly defined, then I think the check for all the cases is quite an easy thing. My problem is, however, whether I can define the homomorphism this way and how to show this is the correct definition.
Thank you very much.
Best Answer
If you mean to take $H_1=H_2$ which swaps $a$ and $b$ (that is, $H_1(a)=b,\ H_1(b)=a$) and fixes all the other elements, then your verification for part 1 works fine. (Though the description of case 2 could be clearer: 'if $x\in K$ or $y\in K$'.)
For part 2, observe that $b=x+x$ for some element $x\in A$ while $a$ cannot be written so.