Proving K(x) is a simple extension of K(u)

Question

I am trying exercises from Algebra by Thomas Hungerford and couldn't solve this particular exercise on page 241.

In the field $K(x)$, let $u=x^3 /(x+1) $. Show that $K(x)$ is a simple extension of $K(u)$ and $[K(x) : K(u)]$?

Elements of $K(x)$ are of form $p(x) /q(x)$ where $p$ and $q$ are polynomials and $q(x) \neq 0$ but how to show that adjoining $1$ element to $K(u)$ will be equal to $K(x)$?

I have no intuition on which result I should use.

Kindly shed some light on this.

Thank you!!

Answer 1

Let's break this down. First, what does it mean to be a simple extension?

An extension $E \subseteq F$ is called simple whenever $F = E(\theta)$ for some element $\theta \in F$. That is, we should be able to get everything in $F$ using only $\theta$ and elements of $E$.

For us, we're working with $K(u) \subseteq K(x)$. By definition we can get everything in $K(x)$ using only $x$ and elements of $K$. So if we allow ourselves $x$ and elements of $K(u)$ (which is bigger than just $K$), we can definitely get all of $K(x)$. This shows us, by taking $\theta = x$ in the definition above, that $K(u)(x) = K(x)$.

(Ok, I guess really we've only shown that $K(u)(x) \supseteq K(x)$. As a quick exercise, you should show the other inclusion).

Now we want to compute $[K(x) : K(u)]$. By our work above, this is $[K(u)(x) : K(u)]$, which looks familiar. Let's recall another result, which is the source of the familiarity:

If $E \subseteq E(\theta)$ is algebraic, then $[E(\theta) : E]$ is the degree of the minimal polynomial of $\theta$ over $E$.

For us, again we have $E = K(u)$, so to compute $[K(u)(x) : K(u)]$ we want to compute the degree of the minimal polynomial of $x$ over $K(u)$. Finally, it's time to use the definition of $u$!

$$u = \frac{x^3}{x+1}$$

We want to write $x$ as the solution to a polynomial in $u$... It seems like the only sensible thing is to try to rearrange this equation and see what happens! We find

$$x^3 - ux - u = 0$$

which makes $x$ a root of the polynomial $p(t) = t^3 - ut - u \in K(u)[t]$. This tells us the minimal polynomial divides $p(t)$. We might expect the minimal polynomial to actually be $p$, and indeed it is. It suffices to show $p$ is irreducible (do you see why?), and as is so often the case, Eisenstein's Criterion comes to the rescue. This is Theorem $6.15$ on page $164$, at least in my copy of Hungerford:

Let $D$ be a unique factorization domain with quotient field $F$. If $f = \sum_{i=0}^n a_i t^i \in D[t]$, with $\text{deg}(f) \geq 1$ and $p$ is an irreducible element of $D$ so that

• $p \not \mid a_n$
• $p \mid a_i$ for $i = 0, 1, 2, \ldots, n-1$
• $p^2 \not \mid a_0$

then $f$ is irreducible in $F[t]$. If $f$ is primitive, then $f$ is irreducible in $D[t]$

Now we notice $u$ is irreducible in $K[u]$ (do you see why?) and moreover

• $u \not \mid 1 = p_3$
• $u \mid p_i$ for $i = 0,1,2$
• $u^2 \not \mid u = p_0$

So by Eisenstein, $p(t)$ is irreducible.

But this is exactly what we wanted! Since $p$ is irreducible, it must be the minimal polynomial of $x$, and so

$$[K(u)(x) : K(u)] = \text{deg}(p) = 3$$

finishing the problem.

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I hope this helps ^_^

Answer 2

Question: "Elements of $K(x)$ are of form $p(x)/q(x)$ where $p$ and $q$ are polynomials and $q(x)≠0$ but how to show that adjoining 1 element to $K(u)$ will be equal to $K(x)$?"

Answer: In general if $k$ is a field and $t$ an independent variable, you may consider any element $u:=f(t)/g(t) \in k(t)$. Assume $(f^n,g^m)=1$ for all integers $n,m\geq 1$. It follows $u$ behaves like an "independent variable" over $k$: The element $u$ does not satisfy any polynomial equation with coefficients in $k$: Assume $F(T):=a_0+\cdots +a_dT^d$ is a polynomial in $k[T]$ with $F(u)=0$. we get an equation

$$a_df^d=-(a_0g^{d-1}+\cdots + a_{d-1}f^{d-1})g$$

and hence $g$ divides $f^d$ a contradiction. Hence the ring $k[u] \subseteq k(t)$ is a polynomial ring.

Example: The element $u:=f(x)/g(x):=x^3/(x+1) \in k(x)$ satisfies $(f^n,g^m)=1$ for all $n,m\geq 1$ hence the ring $k[u] \subseteq k(x)$ is a polynomial ring on the element $u$. Since $k(x)$ is a field it follows $k(u) \subseteq k(x)$ and we get inclusions $k \subseteq k(u) \subseteq k(x)$. Since $x$ satisfies $P(x)=x^3-ux-u=0$, with $P(T):=T^3-uT-u \in k(u)[T]$ is follows $x$ is integral over $k(u)$. Since $k(u)$ is an integral domain it follows the ring $k(u) [x]$ - which is integral over $k(u)$ - is a field and must equal $k(x)$, hence there is an equality $k(x) \cong k(u)[x]$, and we have proved the assertion.