polynomial 1 Attachment(s) I would like to have verification if the following attached proof is correct. If it is not correct, what can be done to make it correct? Thanks. 
You have only shown that a factorization of $f(x)$ gives rise to a factorization of $f(x + c)$ which is trivial. The problem is that EVERY polynomial over a field splits linearly over some extension field. What you haven't shown is that the $h(x+c)$ and $g(x+c)$ lie in $F[x]$ which is required to establish a contradiction. I think the way to prove this is to note that an ideal of $F[x]$ is maximal if and only if it is principally generated by an irreducible (over $F$) polynomial in $F[x]$. This means that $f \in F[x]$ is irreducible is equivalent to the statement that $F[x]/(f)$ is a field. 
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Hi SDK, Thanks for the advice. I have done up as attached. Please see if it is correct. Thanks. 
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I agree with OP. Given f(x+c) is irreducible. Assume f(x) = g(x)h(x) Then f(x+c) =g(x+c)h(x+c), contradiction Note: f(x+c) has same degree as f(x). (ditto g(x+c)) 
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$\displaystyle g(x)=x^{n}+a_{1}x^{n1}+...... $ deg n $\displaystyle g(x+c)=(x+c)^{n}+a_{1}(x+c)^{n1}+......=x^{n}+........$ deg n So f(x+c) is reducible (can be factored into nonconstant polynomials) if f(x) is. But f(x+c) is irreducible. Therefore so is f(x). 
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It is more subtle than that. If you aren't mentioning fields or polynomials rings then you have missed the point. As an example: Is $x^2  2$ irreducible? How about $x^2 + 1$? 
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Write $h(x) = a_0 + a_1 x + \dots + a_n x^n \in F[x]$. Then $h(x+c)$ can be expressed as a sum of monomials of the form $a_m {m\choose k} c^k x^{mk}$ (where, here, ${m\choose k}$ really means the multiplicative identity of $F$ added to itself ${m\choose k}$ times). We know $c$ lies in $F$ and $a_m$ lies in $F$ for all $m$, so each such monomial is in $F[x]$. So any sum of these monomials (and in particular $h(x+c)$) lies in $F[x]$. Same idea for $g$. 
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