The unit circle of a field has an injective homomorphism to the group of units.

Question

Let $K$ be a field with characteristic more than 2, and containing an element $i\in K$ such that $i^2=-1$. Define the unit circle of $K$ to be the set $U.C. = \{(x,y)\in K^2:x^2+y^2=1\}$. Define $\varphi: U.C.\to K^\times$ by

$$ \varphi(x,y) = x+iy $$.

Show that $\varphi$ is injective.

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My work so far: I've already shown that $\varphi$ is a homomorphism. In order to prove injectivity we let $w,x,y,z\in K$ such that $w^2+x^2=1=y^2+z^2$, and need to show that $\varphi(w,x) = \varphi(y,z)$ implies that $(w,x)=(y,z)$. Of course the assumption is the same as

$$ w+ix = y+iz $$

but we can't use the usual manipulation of complex numbers. We can obtain

$$ w-y = i(z-x) $$

which implies

$$ w^2-2wy + y^2 = -(z^2-2xz+x^2) $$

which implies

$$ 2-2wy = 2xz $$

and then

$$ 1 = wy+xz $$

This looks like Bezout's identity and therefore implies a few statements about GCDs. However, none of them implies the equality $w=y$ as far as I can tell.

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Here are the things I know about fields, which seem even possibly relevant:

• Every ideal in $K[x]$ is principal.

• A commutative ring with identity is a field if and only if it has only two ideals.

• An injective homomorphism from an integral domain to a field always extends to an injective homomorphism from its ring of fractions to the same field.

Answer 1

If $i^2 = -1$, then $x^2 + y^2 = (x + iy)(x - iy) = \phi(x, y) (x - iy)$ for any $x$ and $y$. If you are given that $x^2 + y^2 = 1$, you have $x - iy = \frac{1}{\phi(x, y)}$, so that:

\begin{align*} x &= \frac{\phi(x, y) + \frac{1}{\phi(x, y)}}{2} \\ y &= \frac{\phi(x, y) - \frac{1}{\phi(x, y)}}{2i} \end{align*}

So you can recover $x$ and $y$ from $\phi(x, y)$ if $x^2 + y^2 = 1$. This implies that $\phi$ is injective on $\{(x, y) \mid x^2 + y^2 = 1\}$.

Answer 2

Since you already proved that $\varphi : C \to K^{\times}$ is a homomorphism of groups when we equip $K \times K$ and hence $C$ with the "complex number like" multiplication, it suffices to prove that $\varphi$ has trivial kernel, i.e. that $x+iy=1$ implies $x=1$, $y=0$. But $x=1-iy$ means we can substitute it in $1=x^2+y^2$ and get $1=(1-iy)^2+y^2=1-2iy-y^2+y^2=1-2iy$, hence $y=0$ (since $\mathrm{char}(K) \neq 2$) and then $x=1-iy=1$.

Some more background where $\varphi$ is actually coming from (thereby also giving a more conceptual argument why $\varphi$ is a homomorphism): If $K$ is any field, we can construct the $K$-algebra $K[T]/\langle T^2+1 \rangle$, in which the element $[T]$ satisfies $[T]^2+1=0$. It is the universal one: If $R$ is any $K$-algebra with an element $a \in R$ satisfying $a^2+1=0$, there is a unique homomorphism of $K$-algebras $\varphi : K[T]/\langle T^2 + 1 \rangle \to R$ mapping $[T] \mapsto a$. Here, we apply this to $R = K$, $a = i$, and we use the $K$-basis $\{1,[T]\}$ of $K[T]/\langle T^2+1 \rangle$ to identify it with $K \times K$ as a vector space, but the multiplication becomes $(a,c)(b,d)=(ab-cd,ad+bc)$. Of course, the homomorphism $K[T]/\langle T^2+1 \rangle \to K$ will not be injective, but the exercise is about showing that this is the case on a certain subset.