Show that $K( \frac{X^3}{X^2+1} ) ⊂ K(X)$ is an algebraic extension and find the galois group

Question

Let $K$ be a field and $K(X)$ the field of all rational functions over $K$. Show that the expansion $K( \frac{X^3}{X^2+1} )=:L ⊂ K(X)$ is algebraic. Determine its dimension and the associated Galois group.

For showing that $L$ is algebraic over $K(X)$ we must prove tat every element of the extension is a root of a polynomial with coefficients in $K$. Therefore we must find an $f \in K(X)$ with $f(\frac{X^3}{X^2+1}= :a)=0$

I dont know if my idea is right but can we just take $f(t)=t\dot (X^2+1)-X^3$? with this polynomial we have $f(a)=0$

If I just wrote nonsense, how could I solve this problem? Also how do I find the field extension degree and the galois group?

Thank you very much for helping

Answer 1

You just need to show that $X$ is a root of a polynomial with coefficients in $L$. Your choice of $f$ has $f(X^3/(X^2+1))=0$... which just shows $X^3/(X^2+1)$ is a root of a polynomial with coefficients in $K(X)$ but that's the wrong direction! I'll just get you started.

Consider instead the very similar $g(t)=t^3-\frac{X^3}{X^2+1}\cdot t^2-\frac{X^3}{X^2+1}\in L[t]$. Clearly $g$'s evaluation at $X$ in $K(X)$ gives you zero. Is $g$ irreducible in $L[t]$? Fortunately it's a cubic so that amounts to checking if $g$ has any roots. A root $A$ in $L$ would satisfy $A^3=\frac{X^3}{X^2+1}\cdot(A^2+1)$ as an equality in $L$; noting $K[X^3/(X^2+1)]$ is a UFD, we know it is integrally closed - copy the proof of the rational roots theorem - (more generally that’s true because it’s a Dedekind domain) so $A$ would have to be a polynomial. It’s not too hard to check no polynomial $h(x)\in R[x]$, $R$ any commutative ring, can satisfy $h(x)^3=x(h(x)^2+1)$ (hint: think about the lowest order terms).

Ok, so $g$ is irreducible. It follows that $g$ is the minimal polynomial of $X$ over $L$, so that $K(X)/L$ is a degree $3$ algebraic extension. I leave it to you to check that $K(X)/L$ itself is never Galois (it’s always separable but never normal) and thus conclude, for degree reasons, what the Galois group of $g$ is over $L$. Hint: there are two cases we should handle according to whether or not $K$ has characteristic $2$, but in both cases you can use the fact $K[X]$ is integrally closed to deduce the equations you need to solve cannot be solved in $K(X)$. You can start by trying to factorise $g$.

Answer 2

Write $Y = \frac{X^3}{X^2 + 1}$. You seek the minimal polynomial of $X$ over $K(Y)$. Well, $X$ satisfies the polynomial $$ x^3 - Y(x^2 + 1) $$ by construction.

This is a monic polynomial in $K(Y)$, and it is irreducible by Eisenstein's criterion (for example) since the ideal generated by $Y$ in $K[Y]$ is prime, so it is the minimal polynomial.

If I'm reading the Wikipedia right (https://en.wikipedia.org/wiki/Discriminant#Degree_3), this polynomial has discriminant $4Y^3 - 27Y^2$ and since this is not a square in $K(Y)$, the Galois group is $S_3$.