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I have found this question on the Papantonopoulou's Algebra book:

Let $f(x)$ and $g(x)$ be irreducible polynomials over a field $F$ with deg$ f(x) = 15$ and deg$ g(x) = 14$. Let $\alpha$ be a zero of $f(x)$ in some extension field of $F$. Show that $g(x)$ is still irreducible over $F(\alpha)$.

I have some ideas about how to prove it, but none of then use the degrees 15 and 14, which makes me think they will fail...

For instance, I know that a polynomial $p(x)$ is irreducible over a field $K$ iff $\frac{K[x]}{\langle p(x) \rangle}$ is a field. So, I could try to find a isomorphism between $\frac{F(a)[x]}{\langle g(x) \rangle}$ and any field...

I know that the conditions given in the exercise imply in $[F(\alpha) : F ] = 15$, so, maybe I could generate a contradiction by supposing that $g(x)$ is reducible over $F(\alpha)$ and finding basis for $F(\alpha)$ with less than $15$ elements...

Well, anybody has an idea about how to solve this exercise?

Thank you very much.

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  • $\begingroup$ Professor Papantonopoulou, that's Greek to me. $\endgroup$ – Georges Elencwajg Apr 25 '16 at 19:35
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As you noted, $\;[F(\alpha):F]=\deg f=15\;$, so if $\;g(x)\;$ is reducible in $\;F(\alpha)[x]\;$, say $\;g(x)=h(x)k(x)\;,\;\;h,k\in F(\alpha)[x]\;$ and say $\;\beta\;$ is a root of $\;h(x)\;$ , then

$$\;\begin{cases}I\;\;[F(\alpha)(\beta):F(\alpha)]=m<14\\{}\\II\;\;F(\beta)(\alpha):F(\beta)]=m'\le15\end{cases}\;,\;\;\text{yet}\;\;[F(\beta):F]=14\;,\;\;[F(\alpha):F]=15\;,\;\;\text{so}$$

$$[F(\alpha,\beta):F]=\begin{cases}[F(\alpha)(\beta):F(\alpha)][F(\alpha):F]=15m\\{}\\ [F(\beta)(\alpha):F(\beta)][F(\beta):F]=14m'\end{cases}$$

Observe that it must be $\;m=14\cdot x\;,\;\;x\in\Bbb N\;$ , yet this would contradict $I$ above

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  • $\begingroup$ But why $\beta$ must exist? $\endgroup$ – Hilder Vítor Lima Pereira Apr 25 '16 at 20:07
  • $\begingroup$ @Vitor Any polynomial of positive degree over any field has a root in some extension field (or, if you prefer, in an algebraic closure of $\;F\;$ , in this case) $\endgroup$ – DonAntonio Apr 25 '16 at 20:14
  • $\begingroup$ Ok, sure. I thought you were taking $\beta \in F(\alpha)$... $\endgroup$ – Hilder Vítor Lima Pereira Apr 25 '16 at 20:25
  • $\begingroup$ $m$ can't be zero, so $x$ can't be zero, therefore, $m = 14x > 14$, which is a contradiction... It seems correct. I just didn't understand yet how did you find the condition I I. Could you please explain it? $\endgroup$ – Hilder Vítor Lima Pereira Apr 25 '16 at 20:31
  • $\begingroup$ @Vitor The idea came from seeing the polynomials' degrees are coprime. This example shows the condition can be easily generalized to elements which minimal polynomials have coprime degree. $\endgroup$ – DonAntonio Apr 25 '16 at 21:21

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