I was thinking of using contradiction by assuming that that $f(x)$ is reducible but I really don't know how to continue from that idea.
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Hint: can you show that $f(x)=g(x)h(x)$ if and only if $f(x+1)=g(x+1)h(x+1)$? – Jyrki Lahtonen Mar 19 '15 at 18:55
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$g(x)$ is a factor of $f(x)$ if and only if $g(x+1)$ is a factor of $f(x+1)$... – rogerl Mar 19 '15 at 18:55
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Assume that $f(x)$ is reducible. Then:
$\exists \ h(x), \ g(x) \in \mathbb K[x]$ such that $f(x) = g(x)h(x)$.
Then: $f(x + 1) = g(x+1)h(x+1)$, and so reducible. A contradiction.
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Argument is missing an important point, e.g. why can't $,g(x+1),$ be constant? – Bill Dubuque Mar 19 '15 at 18:58
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You need to explicitly mention that, else the proof is incorrect. – Bill Dubuque Mar 19 '15 at 18:59
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I wouldn't even call it an argument, it was just a hint. Details are left to the OP. – Mar 19 '15 at 18:59
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If it is a hint, then you should say that, and mention that some details are left to the reader. – Bill Dubuque Mar 19 '15 at 19:00
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3By the way, the proof is missing, and not "incorrect". Moreover, we are not doing tests here, so it isn't a great deal not to mention one thing or a two. – Mar 19 '15 at 19:09
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Let's prove the contrapositive: That if $f(x)$ is reducible then $g(x) \equiv f(x+1)$ is also reducible. That is logically equivalent to the statement to be proven.
Since $f(x)$ is reducible there exists some $x_0\in K:f(x_0)=0$. Then since $1 \in K$ and $K$ is a field, $y\equiv x_0-1 \in K$. And $$ g(y) = f((x_0-1)+1) = f(x_0) = 0 $$ so $g(x)$ is reducible.
EDITED in response to Bill's comment:
This "proof" is invalid. Yes, if $f(x)$ has a zero then $g(x)$ is irreducible. But we don't know that $f(x)$ has a zero; all we know is that it is irreducible. So "step 1.5" in the above proof is hogwash. Sorry.
Mark Fischler
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2Reducible polynomials need not have roots, e.g. $,(x^2+1)^2 \in\Bbb R[x].\ $ – Bill Dubuque Mar 19 '15 at 19:04
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@Bill Dubuque: You are right, I goofed. Thanks for pointing it out. – Mark Fischler Mar 19 '15 at 22:53