Past exam of the mathematics course of the University of Cambridge / 2008 / ia / Paper 4 / 7D (Numbers and Sets)

Suppose $x^2+y^2=0$ in the field $F$. If $y\ne 0$, division by $y^2$ gives $(x/y{)}^2=-1$, contrary to the hypothesis. Hence $y=0$ and then $x^2=0$; the absence of nonzero zero divisors in a field gives $x=0$. ThereforeFor the stated operations on $F^2$, addition is an abelian group operation: associativity and commutativity hold componentwise, its identity is $(0,0)$, and the negative of $(x,y)$ is $(-x,-y)$.

To check multiplication without omitting associativity, associate to each pair the matrixThis is an injection into the two-by-two matrices over $F$, and direct matrix multiplication givesIt also preserves addition. The associativity and distributivity of matrix multiplication therefore imply the same laws for the pair multiplication. Its formula is symmetric in the pairs, so it is commutative. Its identity is $(1,0)$, which is different from $(0,0)$.

For a nonzero pair, the first result shows that $x^2+y^2\ne 0$. Its multiplicative inverse issince multiplying $(x,y)$ by $(x,-y)$ gives $(x^2+y^2,0)$. This verifies every field axiom, so the given operations make $F^2$ a field. The embedded copy of $F$ is $\{(x,0):x\in F\}$; the element $(0,1)$ has square $(-1,0)$, describing a quadratic extension of $F$.

Suppose $x^2\equiv -1(\mathrm{mod}p)$. Then $x\ne 0$ modulo the prime number $p$, so Fermat's little theorem gives $x^{p-1}\equiv 1(\mathrm{mod}p)$. But $p=4m+3$ makes $(p-1)/2=2m+1$ odd, and thereforeThis contradicts $1\not\equiv -1(\mathrm{mod}p)$, since $p$ is odd. Thus $-1$ is not a quadratic residue modulo $p$. Apply part (a) to the finite field $F={\mathbb{F}}_p$: the pairs $F^2$ with the given operations form a field, and there are $p$ choices for each coordinate, so its cardinality is $\boxed{p^2}$.