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Let $F$ be a field and let $K$ be an extension of $F$. Let $\alpha$ be algebraic over $F$. Let $p(x)$ be the polynomial of minimal degree having $\alpha$ as a root. Prove that $p(x)$ is irreducible in $F[x]$. This is a new concept for me about field extension and i am having difficulty with this and understanding the notation because i kept confusing myself with quotient. I was thinking is it safe to say that $$F[x]/(p(x)) \cong F$$ since $F$ is a field thus $F[x]/(p(x))$ is also a field and also a finite integral domain thus $p(x)$ is maximal, hence prime thus in an integral domain a prime element is always irreducible. Therefore $p(x)$ is irreducible.

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  • $\begingroup$ @Matt Samuel i was wondering if you could check on my reasoning for this problem about what i have said in the post and let me know if that is understandable or not. thanks $\endgroup$ – user146269 May 19 '15 at 18:13
  • $\begingroup$ It's false that the quotient is isomorphic to $F$. It's true that it's a field, but we don't know that because that is what we are trying to prove. $\endgroup$ – Matt Samuel May 19 '15 at 18:17
  • $\begingroup$ oh ok thanks that was the only approach i had in this problem. Well thanks for your help i truly appreciate it. $\endgroup$ – user146269 May 19 '15 at 18:18
  • $\begingroup$ Good enough answers have been posted already so I do not see the point in answering your question but have something to point out. Make sure you accept answers to show appreciation to the answer. (I saw that you have not accepted any answers in any of your question so far) $\endgroup$ – Jack Yoon May 19 '15 at 18:26
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Hints:

== $\;p(x)\;$ is irreducible $\;\implies p(x)=g(x)h(x)\implies g(x)\,\,or\,\,h(x)\;$ is constant

== $\;p(x)\;$ is the minimal polynomial of $\;\alpha\;$ over $\;\Bbb F\implies\;\forall\;h(x)\in\Bbb F[x]\;$ , then $\;h(\alpha)=0\iff p(x)\mid h(x)\;$ .

Further hint for the second hint: divide with residue (Euclidean Algorithm) $\;h(x)\;$ by $\;p(x)\;$ and apply the condition $\;h(\alpha)=0\;$ and minimality of $\;p(x)\;$.

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Suppose $p(x)=q(x)r(x)$ with both $q$ and $r$ of positive degree, necessarily strictly smaller than the degree of $p$. Then $q(\alpha)r(\alpha)=0$, so either $q(\alpha)=0$ or $r(\alpha)=0$, contradicting the assumption that the degree was minimal.

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