rational numbers field axioms Let $\mathbb Q$ be the rational number field.
Is the group $K=\left\{\left.\begin{pmatrix}
a & 2b\\ 
b & a
\end{pmatrix}~\right|~ a,b\in \mathbb Q\right\}$ a field with the regular addition and Scalar multiplication operations?
 A: If the operations are matrix addition and matrix multiplication, the answer is yes, $K$ is a field. If the multiplication is scalar multiplication, then we have not defined the product of elements of $K$.
To prove that under matrix addition and multiplication, $K$ is a field, we need to show that the product of two matrices of the given shape is a matrix of the given shape. To do this,  we  compute
$$\begin{pmatrix}a_1&2b_1\\b_1&a_1\end{pmatrix}\begin{pmatrix}a_2&2b_2\\b_2&a_2\end{pmatrix}$$
and verify it has the right shape.
We also need to show that if $M$ is a non-zero matrix of the right shape, then $M^{-1}$ exists and has the right shape. 
For existence, we show that if $a$ and $b$ are not both $0$, then the determinant of our matrix is non-zero. To do this we need to verify that $a^2-2b^2\ne 0$. That is true because of the irrationality of $\sqrt{2}$.
Finally, the usual formula for the inverse of a $2\times 2$ matrix shows that the inverse is of the right shape. For the inverse of the generic non-zero matrix in our collection is
$$\frac{1}{a^2-2b^2}\begin{pmatrix}a&-2b\\-b&a\end{pmatrix}.$$
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A: In general, given a square matrix of rational numbers $W,$ say $n$ by $n,$ then we can form the commutative ring of all
$$ a_0 I + a_1 W + a_2 W^2 + \cdots a_{n-1}W^{n-1}.   $$
It is not necessary to use higher powers because of the Cayley-Hamilton theorem. If, in addition, the characteristic polynomial $f(x)$ of $W$ is irreducible over $\mathbb Q,$ we not only have a lack of zero divisors, we also have a field  isomorphic to
$$ \mathbb Q[x]/ (f(x)) $$
A: Recall that for a field $(K,+,\times)$ we need to have two operations $+$ and $\times$ which satisfy the field axioms.  You define $K$ to be a particular set of $2 \times 2$ matrices.  But scalar multiplication (ie, multiplying a scalar $c \in \mathbb{Q}$ with a $2 \times 2$ matrix $k \in K$) does not give a well defined system $(K, +, \times)$ since it does not define the binary operation $\times$ for elements in $K$.  So, the answer to your question is a negative. 
(Scalar multiplication arises in the axioms of some other algebraic structure - a vector space, for example.)
