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Let $f(x)$ and $g(x)$ be irreducible polynomials over a field $F$ and let $a,b \in E$ where $E$ is some extension of $F$. If $a$ is a zero of $f(x)$ and $b$ is a zero of $g(x)$, show that $f(x)$ is irreducible over $F(b)$ if and only if $g(x)$ is irreducible over $F(a)$.

Attempt: Since $f(x),g(x)$ are irreducible over $F \implies a,b \notin F$.

$f(x)$ is irreducible over $F(b) $ and $f(x)$ is irreducible over $F \implies a \neq b$ (As, $a$ is the zero of $f(x)$)

Which means $b \notin F(a)$ either $\implies g(x)$ is irreducible over $F(a)$.

Similarly, the other half can be proved in a similar way.

Is my solution attempt correct?

Thank you for your help..

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  • $\begingroup$ Since, $a$ is a zero of $f(x)$ and $f(x)$ is irreducible over $F(b) \implies f$ does not have a zero in $F(b). F(b)$ is the smallest field containing $F$ and $b$. Since, $f$ does not have a zero in $F(b)$.Hence, we can be sure that $a \neq b$ else $f$ would have a zero in $F(b)$. $\endgroup$ – MathMan Aug 17 '14 at 10:57
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    $\begingroup$ @VHP The problem is that "$f$ does not have a zero in $F(b)$" is a necessary, but not sufficient, condition for $f$ to be irreducible in $F(b)$. $f$ could factor into two irreducible quadratics, for example. So you really need to draw a stronger conclusion here. See user26857's correct answer. $\endgroup$ – Slade Aug 17 '14 at 16:18
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$[F(a,b):F(b)]=\deg f$ iff $f$ is irreducible over $F(b)$. In this case $[F(a,b):F]=\deg f\deg g$.

$[F(a,b):F(a)]=\deg g$ iff $g$ is irreducible over $F(a)$. In this case $[F(a,b):F]=\deg f\deg g$.

We also have $[F(a,b):F]=[F(a,b):F(b)][F(b):F]=(\deg g)[F(a,b):F(b)]$ and $[F(a,b):F]=[F(a,b):F(a)][F(a):F]=(\deg f)[F(a,b):F(b)]$.

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    $\begingroup$ Could you explain why $[F(a,b) : F(b)] =$deg $f$ ($[F(a,b) : F(a)] =$deg $g$ , respectively) proves that g is irreducible over $F(a)$ (f is irreducible over $F(b)$ , respectively)? $\endgroup$ – user337254 Oct 17 '18 at 15:00
  • $\begingroup$ Could you please explain that point? $\endgroup$ – KingDingeling Mar 14 '19 at 16:14
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    $\begingroup$ $[K(\alpha):K]$ equals the degree of the minimal polynomial of $\alpha$ over $K$. $\endgroup$ – user26857 Mar 15 '19 at 10:10
  • $\begingroup$ thank you for your help! $\endgroup$ – KingDingeling Mar 15 '19 at 21:53
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Hint $\qquad \begin{array}{ccc} & F(\alpha,\beta)\ &\\ \color{#c00}x\nearrow\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!& &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \nwarrow \color{#0a0}y\\ F(\alpha)\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! & &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! F(\beta)\\ & a\nwarrow\qquad\nearrow b \\ & F & \end{array} \Rightarrow\ \ \ {xa = yb}\ \ \ \Rightarrow\!\!\!\!\!\! \overset{\Large \stackrel{g\ {\rm irred\ over\ } F(\alpha)\ \ \ \ \ }\Updownarrow}{\color{#c00}{x=b}}\!\!\!\!\!\!\!\!\!\iff\!\!\!\!\!\!\!\!\! \overset{\Large \stackrel{\ \ f\ {\rm irred\ over\ }F(\beta)}\Updownarrow_\phantom{I^{I^I}}\!\!\!\!\!\!\!\!}{\color{#0a0}{y = a}}$

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