# Math Help - Proving Irreducibility

1. ## Proving Irreducibility

Let $k$ be a field, let $K=k(t)$ be the function field of $k$.
Let $\alpha \in K$ and set $\alpha = \tfrac{f(t)}{g(t)}$ in reduced terms, with $\alpha \not \in k$.
Now construct $F = k(\alpha)$.
Note $t\in k(t)$ solves the polynomial $\alpha f(x) - g(x)\in F[x]$.
This is a non-zero polynomial because otherwise it would contradict $\alpha \not \in k$.
How do we prove this polynomial is irreducible over $F$?

2. Originally Posted by ThePerfectHacker
Let $k$ be a field, let $K=k(t)$ be the function field of $k$.
Let $\alpha \in K$ and set $\alpha = \tfrac{f(t)}{g(t)}$ in reduced terms, with $\alpha \not \in k$.
Now construct $F = k(\alpha)$.
Note $t\in k(t)$ solves the polynomial $\alpha f(x) - g(x)\in F[x]$.
This is a non-zero polynomial because otherwise it would contradict $\alpha \not \in k$.
How do we prove this polynomial is irreducible over $F$?
since $k(\alpha)$ is the field of fractions of $k[\alpha]$, by Gauss's lemma we only need to prove that $\alpha f - g$ is irreducible over $k[\alpha].$

suppose that $\alpha f(x) - g(x)=a(x)b(x),$ for some $a(x) \in k[\alpha][x], \ b(x) \in k[\alpha][x].$ now look at $a(x), \ b(x)$ as elements of

$k[x][\alpha].$ see that $\alpha$ is transcendental over $k(x).$ now since, as an element of $k[x][\alpha],$ the degree of $\alpha f(x)- g(x)$ is 1, we

must have $a(x)=\alpha u(x) + v(x),$ and $b(x)=w(x),$ for some $u(x),\ v(x),$ and $w(x)$ in $k[x].$ therefore $u(x)w(x)=f(x)$ &

$v(x)w(x)=-g(x).$ hence $w(t)$ must divide $\gcd(f(t),g(t))=1.$ so $w(t) \in k^{\times}.$ hence $b(x)=w(x) \in k^{\times}. \ \ \square$

3. Originally Posted by NonCommAlg
$u(x)w(x)=f(x)$ &

$v(x)w(x)=-g(x).$
I think you mean, $\alpha f(x) = u(x)w(x)$, but your argument still works. Thank you.

It seems to me that the transcendental property of $\alpha$ was very important over here. And I am guessing that is because: the ring generated by a transcendental element is isomorphic to the polynomial ring in a naturally manner, i.e. $k[\alpha]\simeq k[x]$ by sending $\alpha\mapsto x$ and building the isomorphism out of that. The crucial point is that this mapping is well-defined because if an element of $k[\alpha]$ had two different representations in terms of $\alpha$ then that would contradict the fact that $\alpha$ is transcendental. Am I getting the right idea?

Off topic question: why did you choose to study non-commutative algebra over field theory? Is it because it is more general?

4. Originally Posted by ThePerfectHacker
I think you mean, $\alpha f(x) = u(x)w(x)$, but your argument still works. Thank you.
No! the correct one is what i wrote, i.e. $f(x)=u(x)w(x).$ the reason is that from $\alpha f(x)-g(x)=(\alpha u(x) + v(x))w(x)$

we'll get $(f(x)-u(x)w(x))\alpha -g(x)-w(x)=0.$ but since $\alpha \notin k(x),$ we get $f(x)-u(x)w(x)=-g(x)-w(x)=0.$

Off topic question: why did you choose to study non-commutative algebra over field theory? Is it because it is more general?
it's actually noncommutative ring theory not only algebras. why did i choose it? i guess because i think ring theory is interesting

and the theory of commutative rings is too old and, well, too easy!

5. Originally Posted by NonCommAlg
it's actually noncommutative ring theory not only algebras. why did i choose it? i guess because i think ring theory is interesting
I forgot to ask, since this is related to what you do, is there such a thing as Galois theory over skew fields? More specifically let $F\subseteq K$ be skew fields, is there a way to determine all skew fields $E$ so that $F\leq E\leq K$?

My other question is about the formal meaning of $k[x][\alpha]$. I understand it to mean as all polynomials in $\alpha$ with coefficients in $k[x]$, and this is okay because of transcendentalness. My problem is how do we formally define $k[x][\alpha]$? I never seen it used before.