= Solution
Suppose $x^2+y^2=0$ in the <field> $F$. If $y\ne0$, 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 $x=0$. Therefore
$$
\boxed{x^2+y^2=0\Longrightarrow x=y=0.}
$$
For 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 <matrix>
$$
M(x,y)=\begin{pmatrix}x&-y\\y&x\end{pmatrix}.
$$
This is an <injection> into the two-by-two <matrices> over $F$, and direct <matrix multiplication> gives
$$
M(x,y)M(z,w)=M(xz-yw,xw+yz).
$$
It 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 $x^2+y^2\ne0$. Its multiplicative inverse is
$$
\boxed{(x,y)^{-1}=\left(\frac{x}{x^2+y^2},\frac{-y}{x^2+y^2}\right),}
$$
since multiplying $(x,y)$ by $(x,-y)$ gives $(x^2+y^2,0)$. This verifies every <field> axiom, so \b[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 $F$.
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