HW15

Let ${\mathbb{F}}_p$ be the finite field of $p$ elements.

• When $p=2,3,5,7$, show that $\forall x\in {\mathbb{F}}_p$, there exists some $a,b\in {\mathbb{F}}_p$ (including the zero element), such that $x=a^2+b^2$.
• Show that the above assertion is true for any $p$.

\begin{proof} It suffices to show the second argument. Assume that the characteristic of ${\mathbb{F}}_p$ is $p_0$.

When $p_0=2$, note that the Frobenius map $\varphi :a\to a^2$ is an field automorphism. Then for any $x\in {\mathbb{F}}_p$, there exists $y\in {\mathbb{F}}_p$ such that $\varphi (y)=y^2=x$ and so $x=y^2+0^2$.

Now consider the case of $p_0>2$. Since ${\mathbb{F}}_p^{\times }$ is a cyclic group of order $p-1$, there exists $\lambda \in {\mathbb{F}}_p$ such that ${\mathbb{F}}_p^{\times }=⟨\lambda ⟩$ and so $\{{\lambda }^{2k}:0<2k⩽p-1\}$ is a set of square elements. In addition, $0=0^2$ is also a square element, thus there are $(p+1)/2$ elements of square elements. For any $x\in {\mathbb{F}}_p$, define

$$A:=\{a^2:a\in {\mathbb{F}}_p\}\text{mbox}andB:=\{x-b^2:b\in {\mathbb{F}}_p\}.$$

By the argument above, there is $|A|=|B|=(p+1)/2$. Since $|A|+|B|>p$, there exists $m\in A\cap B$ where $m=a_m^2=x-b_m^2$. It deduces that $x=a_m^2+b_m^2$. By the arbitrary of $x$, we finish the proof. \end{proof}

In the following exercises, $K(x)$ means the fractional field of $K[x]$ (here $K[x]$ is the polynomial ring with a transcendental variable $x$ ).

6. Let $f/g\in K(x)$ with $f/g\notin K$ and $f,g$ relatively prime in $K[x]$ and consider the extension of $K$ by $K(x)$.

• (a) $x$ is algebraic over $K(f/g)$ and $[K(x):K(f/g)]=\max (\mathrm{deg}f,\mathrm{deg}g)$. (Hint: $x$ is a root of the nonzero polynomial $\phi (y)=(f/g)g(y)-f(y)\in K(f/g)[y]$; show that $\phi$ has degree $\max (\mathrm{deg}f,\mathrm{deg}g)$. Show that $\phi$ is irreducible as follows: Since $f/g$ is transcendental over $K$ (why?) we may for convenience replace $K(f/g)$ by $K(z)$ ($z$ an indeterminate) and consider $\phi =zg(y)-f(y)\in K(z)[y]$. By Lemma III. $\phi$ is irreducible in $K(z)[y]$ provided it is irreducible in $K[z][y]$. The truth of this latter condition follows from the fact that $\phi$ is linear in $z$ and $f,g$ are relatively prime.)
• (b) If $E\ne K$ is an intermediate field, then $[K(x):E]$ is finite.
• (c) The assignment $x\mapsto f/g$ induces a homomorphism $\sigma :K(x)\to K(x)$ such that $\phi (x)/\psi (x)\mapsto \phi (f/g)/\psi (f/g)$. $\sigma$ is a $K$ automorphism of $K(x)$ if and only if $\max (\mathrm{deg}f,\mathrm{deg}g)=1$.
• (d) ${\mathrm{Aut}}_{K}K(x)$ consists of all those automorphisms induced (as in (c)) by the assignment $x\mapsto (ax+b)/(cx+d)$, where $a,b,c,d\in K$ and $ad-bc\ne 0$.

\begin{proof} a) To show $x$ is algebraic over $K(f/g)$, it suffices to show there exists $\phi (y)\in K(f/g)[y]$ such that $\phi (x)=0$. Define $\phi (y)=(f(x)/g(x))g(y)-f(y)$, then $\phi (x)=0$ and so $x$ is algebraic. Notice that $\mathrm{deg}\phi =\max (\mathrm{deg}f,\mathrm{deg}g)$, thus to show $[K(x):K(f/g)]=\max (\mathrm{deg}f,\mathrm{deg}g)$ it remains to show $h$ is irreducible in $K(f/g)[y]$.

If $f/g$ is algebraic over $K$, then there exists a polynomial $h\in K[x]$ with $\mathrm{deg}h=m$ such that $h(f/g)=0$. It deduces that $x$ is a root of $\ell (x):=g^m(x)h(f(x)/g(x))\in K[x]$, which contradicts with $x$ transcendental. Therefore, $f/g$ is transcendental over $K$. By Lemma 6.13 of Hungerford, to show $\phi$ is irreducible in $K(z)[y]$, it suffices to show $\phi$ is irreducible in $K[z][y]$.

Assume that $\phi$ is reducible in $K[z][y]$, then there exist ${\psi }_1,{\psi }_2\in K[z][y]$ such that ${\psi }_1{\psi }_2=\phi$. Since $\phi$ is linear in $z$, WLOG we assume ${\psi }_1\in K[y]$ and ${\psi }_2={\psi }_{21}z+{\psi }_{22}$ where ${\psi }_{21},{\psi }_{22}\in K[y]$. It deduces that ${\psi }_1{\psi }_{21}=g$ and ${\psi }_1{\psi }_{22}=-f$, which is impossible as $(f,g)={\psi }_1$. Therefore, $\phi$ is irreducible in $K[z][y]$. Now we finish the proof.

b) Take $f_1/g_1\in E\setminus K$, then by a) $[K(x):K(f_1/g_1)]<\infty$. Since $E\supseteq K(f_1/g_1)$, we have $[K(x):E]⩽[K(x):K(f_1/g_1)]<\infty$.

c) Note that $\sigma {|}_K=\mathrm{id}$. If $\sigma$ is a $K$-automorphism, then $K(x)=\sigma (K(x))\subseteq K(f/g)\subseteq K(x)$ and so $K(x)=K(f/g)$. By a), we have $1=[K(f/g):K(x)]=\max (\mathrm{deg}f,\mathrm{deg}g)$. Conversely, if $\max (\mathrm{deg}f,\mathrm{deg}g)=1$, then $K(x)=K(f/g)$ by a). It is easy to verify $\sigma (K(x))=K(f/g)$, then $\sigma (K(x))=K(f/g)=K(x)$ and so $\sigma$ is an $K$-automorphism.

d) For any $\sigma \in {\mathrm{Aut}}_K(K(x))$, $\sigma (x)=f(x)/g(x)$ and so $\sigma$ is defined as $\phi (x)/\psi (x)\mapsto \phi (f/g)/\psi (f/g)$. By c), if $\sigma$ is a $K$-automorphism, then $\max (\mathrm{deg}f,\mathrm{deg}g)=1$. If one of $f,g$ has degree $1$, then $f,g$ are not relatively prime in $K[x]$. Thus $\mathrm{deg}f=\mathrm{deg}g=1$. Assume that $f=ax+b$ and $g=cx+d$. Since they are relatively prime, there is $b/a\ne d/c$ and $ad-bc\ne 0$. Therefore, ${\mathrm{Aut}}_{K}K(x)$ consists of all those automorphisms induced by the assignment $x\mapsto (ax+b)/(cx+d)$, where $a,b,c,d\in K$ and $ad-bc\ne 0$. \end{proof}

Hungerford GTM Algebra.pdf