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Let $p(x)$ be a minimal polynomial for $a$ for field $F$. This implies it is a monic polynomial of the lowest degree possible such that $p(x)=0$.

Why does $p(x)$ have to be irreducible? Why can't it be split into factors?

If $p(x)$ is reducible, which I think should be the case, then it can have some of its factors in common with other polynomials. Hence, $F[a]$ won't be a field then, in spite of $a$ being algebraic over $F$, because the polynomials which have non-zero factors in common with $p(x)$ will not have a multiplicative inverse.

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if p(x) is reducible, suppose p(x)=f(x)g(x), then f(a)=0 or g(a)=0, but f and g both has lower degree than p, which is not possible, because p has lowest degree.

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  • $\begingroup$ damn. i knew that! thanks. stupid question. $\endgroup$
    – user67803
    May 22, 2013 at 12:51
  • $\begingroup$ The key point is $\,f(a),g(a)\in E\,$ an extension field, so $\,f(a)g(a) = 0\,$ $\Rightarrow$ $\,f(a)=0\,$ or $\,g(a)=0,\ $ i.e. a field is a domain so has no zero-divisors, i.e. nonzero elements are cancellable. This may fail if $\,E\,$ is an extension ring which is not a domain, e.g. $\,t\in R = F[t]/(t^2)\,$ has reducible minimal polynomial $\,x^2,\,$ indeed, $\,t\,$ is a zero-divisor in $\,R,\,$ being nilpotent $\,t^2 = 0.$ This is familiar from linear algebra, where one meets many reducible minimal polynomials of matrices. $\endgroup$
    – Key Ideas
    May 22, 2013 at 15:26

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