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Let $K$ be an extension field of $F$, and let $\alpha \in K$ be algebraic. Suppose that $f(x) \in F[x]$. Prove that $f(x)$ is the unique monic polynomial of least degree with f($\alpha$) = $0$.

My thinking is that if $f(x)$ isn't unique, then there exists another polynomial of $g(x)$ of the same degree. And, if that's the case, then these two polynomials would divide each other, which means $f(x)$ is not irreducible. Is this correct? Or am I missing pieces, or completely misunderstanding things? Thank you for your help.

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  • $\begingroup$ Hint $\ $ If $\,f(a) = 0 = g(a)\,$ and $\,f\ne g\,$ then $\,a\,$ is also a root of their difference, a nonzero lower degree polynomial (lower since both are monic of equal degree, so lead terms cancel). $\endgroup$ – Bill Dubuque Nov 30 '16 at 3:31
  • $\begingroup$ @BillDubuque But if $f(x)$ and $g(x)$ are supposedly the lowest degree polynomials, then no lower degree polynomial can exist, right? Is that the contradiction? $\endgroup$ – Max Nov 30 '16 at 3:35
  • $\begingroup$ Exactly. That contradiction concludes the proof. $\endgroup$ – Bill Dubuque Nov 30 '16 at 3:43
  • $\begingroup$ Oh, just answer my own question? I can do that? (sorry, new to MSE) $\endgroup$ – Max Nov 30 '16 at 3:47
  • $\begingroup$ Yes, and it means you will get feedback (e.g. I corrected your division to a difference). $\endgroup$ – Bill Dubuque Nov 30 '16 at 3:55
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Either $f(x)$ divides $g(x)$ or the vice versa .

Let WLOG;$g(x)|f(x)\implies f(x)=g(x)q(x)+r(x)$ where $\deg r(x)<\deg g(x)$...

$f(a)=0\implies r(a)=0$ which is false as $\deg r(x)<\deg g(x)\le\deg f(x)$

Hence $r=0\implies f(x)=cg(x);c\text{is a constant}$

But that is false as $f,g$ are monic.So $f=g$

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Suppose $f(x)$ isn't unique. Then, there exists another monic polynomial of least degree $g(x)$ such that $g(\alpha)$ = $0$.

$\implies f(\alpha) = 0 = g(\alpha)$.

Since $f \ne g$,

$\implies \alpha$ is also a root of a lower degree polynomial $h = $ $f - g$;

($h$ is nonzero and of lower degree since $f$ and $g$ are different, monic nonzero polynomials. So, we have cancelling leading terms, and $h$ consists of the difference of the remaining terms, which are not equal since $f \ne g$, and therefore nonzero).

however, since $f$ and $g$ are of least degree, there cannot exist a polynomial $h$ of lower degree.

$\therefore g$ cannot exist, and $f$ is unique.

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  • $\begingroup$ From the discussion above. $\endgroup$ – Max Nov 30 '16 at 3:53
  • $\begingroup$ You should explicitly justify why $h$ has lower degree (and is nonzero). $\endgroup$ – Bill Dubuque Nov 30 '16 at 3:57
  • $\begingroup$ Alright, I'll add in why I think that is. $\endgroup$ – Max Nov 30 '16 at 3:59
  • $\begingroup$ Note that the difference needn't be monic, so is not immediately a counterexample, but it does yield a monic since .... $\endgroup$ – Bill Dubuque Nov 30 '16 at 4:02
  • $\begingroup$ Is it because $f$ and $g$ are of lowest degree, so the difference would just be a constant? $\endgroup$ – Max Nov 30 '16 at 4:07

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