# Give an example, with justification, to show that: 1.the union of integral domains need not be an integral domain; 2.all finite fields are not isomorphic.?

Feb 9, 2018

See explanation...

#### Explanation:

Example of two integral domains whose union is not one

A rich source of examples of such things is matrices.

For example, note that:

${\left(\begin{matrix}0 & 1 \\ 2 & 0\end{matrix}\right)}^{2} = \left(\begin{matrix}0 & 1 \\ 2 & 0\end{matrix}\right) \left(\begin{matrix}0 & 1 \\ 2 & 0\end{matrix}\right) = \left(\begin{matrix}2 & 0 \\ 0 & 2\end{matrix}\right)$

That is $\left(\begin{matrix}0 & 1 \\ 2 & 0\end{matrix}\right)$ is a square root of $2$.

We can identify an integral domain of matrices of the form:

$m \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) + n \left(\begin{matrix}0 & 2 \\ 1 & 0\end{matrix}\right) = \left(\begin{matrix}m & n \\ 2 n & m\end{matrix}\right)$

with $m , n \in \mathbb{Z}$ which is isomorphic to the ring of numbers of the form $m + n \sqrt{2}$.

Note that this has no zero divisors since $\sqrt{2}$ is irrational.

Similarly, we find:

${\left(\begin{matrix}0 & 2 \\ 1 & 0\end{matrix}\right)}^{2} = \left(\begin{matrix}2 & 0 \\ 0 & 2\end{matrix}\right)$

So $\left(\begin{matrix}0 & 2 \\ 1 & 0\end{matrix}\right)$ is also a square root of $2$ and we can identify an integral domain of matrices of the form

$\left(\begin{matrix}m & 2 n \\ n & m\end{matrix}\right)$

which is also isomorphic to the ring of numbers of the form $m + n \sqrt{2}$

The union of these two sets of matrices is not closed under addition and therefore not a ring, let alone an integral domain.

For example:

$\left(\begin{matrix}0 & 1 \\ 2 & 0\end{matrix}\right) + \left(\begin{matrix}0 & 2 \\ 1 & 0\end{matrix}\right) = \left(\begin{matrix}0 & 3 \\ 3 & 0\end{matrix}\right)$

is not in either of the two integral domains.

Furthermore, if we add all elements required to make the set of matrices closed under addition and additive inverse, then we get a ring with zero divisors. For example:

$\left(\left(\begin{matrix}0 & 3 \\ 3 & 0\end{matrix}\right) - \left(\begin{matrix}3 & 0 \\ 0 & 3\end{matrix}\right)\right) \left(\left(\begin{matrix}0 & 3 \\ 3 & 0\end{matrix}\right) + \left(\begin{matrix}3 & 0 \\ 0 & 3\end{matrix}\right)\right) = \left(\begin{matrix}- 3 & 3 \\ 3 & - 3\end{matrix}\right) \left(\begin{matrix}3 & 3 \\ 3 & 3\end{matrix}\right) = \left(\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right)$

Finite fields of different orders or characteristics cannot be isomorphic

For example $G F \left(2\right)$ and $G F \left(3\right)$ have $2$ and $3$ elements respectively, so there is not even a bijection between them as sets, let alone an isomorphism of fields.

All finite fields have ${p}^{n}$ elements for some prime $p$ and positive integer $n$. All finite fields of the same order are isomorphic. That is, a finite field is essentially determined by the values of $p$ and $n$.