Prove it is a field and find the number of its elements

Question

Let $P = \left \{\begin{pmatrix} a & \bar{8}b\\ \bar{5}b & a \end{pmatrix} \;\middle|\; a, b \in \mathbb Z_{19}\right \}$.

Prove that $P$ and $\mathbb Z_{19}[x] / (x^2 - \bar{2})$ are fields and find the number of their elements.

How do I do that?

Answer 1

Here are some hints:

1) $P$ is a subset of $M_2(\mathbb{Z}/19 \mathbb{Z})$, the ring of $2 \times 2$-matrices with entries in $\mathbb{Z}/19 \mathbb{Z}$. That means: You don't have to verify that addition and multiplication are associative or that addition is commutative or the distributivity laws. You just have to prove that:

(i) The sum of two elements of $P$ is an element of $P$ again.

(ii) The neutral element of addition of $M_2(\mathbb{Z}/19 \mathbb{Z})$ lies in $P$. (How to choose a,b?)

(iii) For each element $A \in \mathbb{P}$ we find an element $\bar A \in P$ such that $A+ \bar A$ is equal to the neutral element of addition.

(iv) The product of two elements of $P$ is an element of $P$ again.

(v) The neutral element of multiplication of $M_2(\mathbb{Z}/19 \mathbb{Z})$ lies in $P$. (How to choose a,b?)

(vi) Each element in $P$, except the neutral element of addition, is invertible in $M_2(\mathbb{Z}/19 \mathbb{Z})$. From linear algebra you know that you have to examine the determinant to show that. Linear Algebra also offers formulas to calculate the inverse matrix. So the last step is to show that the inverse matrix lies in $P$ again.

Counting the elements of $P$ is quite easy in this case. How many different numbers can $a$ be? How many choices do we have for $b$? What can we hence say about $|P|$?

2) We know that there exists a ring homomorphism $\phi \colon \mathbb{Z}/19 \mathbb{Z}[X] \to P$ which maps $\overline 1$ to $[ \overline 1, \overline 0; \overline 0, \overline 1] \in P$ and $X$ to $[ \overline 0, \overline 8; \overline 5, \overline 0] \in P$. If you are able to prove that $\phi$ is surjective and that $\ker (\phi) = (x^2-\overline 2) \cdot \mathbb{Z}/19 \mathbb{Z}[X]$ (= set of polynomials which are multiples of $x^2-\overline 2$), you receive $P \cong \mathbb{Z}/19 \mathbb{Z}[X]/(x^2-\overline 2)$. That means: $\mathbb{Z}/19 \mathbb{Z}[X]/(x^2-\overline 2)$ is a field, hence $P$ is a field, and both have the same number of elements.

EDIT: You also could verify that $\mathbb{Z}/19 \mathbb{Z}[X]/(x^2-\overline 2)$ is a field ($\Longleftrightarrow x^2-\overline 2$ is irreducible) first. Then the isomorphism implies that $P$ is a field. But counting the elements of $P$ seems to be easier that counting the elements of $\mathbb{Z}/19 \mathbb{Z}[X]/(x^2-\overline 2)$.

Answer 2

I would solve it as follows:

1. Note that $X^2-2$ is irreducible modulo $19$ since $2$ is non-square residue; hence, $F_{19^2} \cong Z_{19}[X]/X^2-2$ is a field. (It has $19^2$ elements since it is a vector space of dimension $2$ over $Z_{19}$, or more down to earth, the elements of $Z_{19}[X]/X^2-2$ are $aX + b\mod X^2-2$, with $a,b\in Z_{19}$.

2. construct isomorphism between $P$ and $F_{19^2}$. For this, first simultaneously diagonalize the elements of $P$ over $F_{19^2}$, and then the isomorphism is given by sending a matrix to its first eigenvalue.