Let $K=F(\frac{x^7}{x^5+x^3+3x+1})$, where $F$ is a field. Determine $[F(x): K]$.

Question

(a) Let $R$ be a unique factorization domain and consider $$ f(x,y) = x^7 + yx^5 + yx^3 + 3yx + y \in R[x,y]. $$ Show that $f(x,y)$ is irreducible.

Proof: Note $R[x,y]=R[y][x]$. Let $S=R[y]$. We will show $f(x,y)$ is irreducible in $S[x]$. By Eisenstein's Criterion, let $p=y$,

• $y | a_0=y, a_1=3y, a_2=0, a-3=y, a_4=0, a_5=y, a_6=0$
• $y \nmid a_7=1$
• $y^2 \nmid a_0=y$

Why is $y$ a prime in $S$? Notice that $S/(y)=R[y]/(y)\cong R$ and $R$ is integral domain. So $(y)$ is a prime ideal in $S$, and hence $Y$ is a prime element in $S$.

Q.E.D.

(b) Let $K=F(\frac{x^7}{x^5+x^3+3x+1})$, where $F$ is a field. Determine $[F(x): K]$.

I think that I need to first show whether $x^5+x^3+3x+1$ is irreducible or not. But I'm not sure if that is right or how to show that or how to follow from there. Can I get some help?

Answer 1

Let $y=\frac{x^7}{x^5+x^3+3x+1}$, this is an element of $K$. We get:

$x^7-yx^5-yx^3-3yx-y=0$

Just like before, you can show that the polynomial $t^7-yt^5-yt^3-3yt-y\in K[t]=F(y)[t]$ is irreducible over $K=F(y)$, and so it is the minimal polynomial of $x$ over this field. So $[F(x):K]=[F(y)(x):F(y)]=7$.