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I need to show that the polynomial $f(x)=x^6+x^5+x^4+x^3+x^2+x+1$ is irreducible in $\mathbb{Z}[X]$ and in $\mathbb{F}_2[X]$. As we can't find a prime number p satisfying the conditions for the Eisenstein criterion, I did not know how to solve it. I looked into the solutions and they apply the Eisenstein criterion to $f(x+1)$ instead of $f(x)$. I don't understand why we can do this.
Could somebody explain this to me? And is proving irreducibility for $f(x+1)$ enough?

Aryaman Maithani
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Mikalo
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  • It's because $f(x+1)$ is an Eisenstein polynomial, while $f(x)$ isn't. And are you trying to prove irreducibility over $\Bbb F_2$? – Angina Seng Jun 17 '20 at 19:35

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I looked into the solutions and they apply the Eisenstein criterion to $f(x+1)$ instead of $f(x)$. I don't understand why we can do this.

It suffices to show irreducibility of $f(x+1)$. To see this, assume that $f(x)$ is reducible, then $f(x) = g(x)h(x)$ for some proper factors $g(x)$ and $h(x)$. In that case, you get that $$f(x+1) = g(x+1)h(x+1)$$ where the right factors are still proper factors. Thus, reducibility of $f(x)$ implies that of $f(x+1)$. The contrapositive shows that it's sufficient to prove the irreducibility of $f(x)$.

Aryaman Maithani
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Irreducibility is the same as lack of rational (or perhaps real) roots. The graph of $f(x+1)$ is simply the graph of $f(x)$, shifted horizontally by one unit. This move preserves the lack (or presence) of rational (or real) roots.

G Tony Jacobs
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  • "Irreducibility is the same as lack of rational (or perhaps real) roots." I may be mis-interpreting the statement but isn't a polynomial like $x^4 + 1 \in \Bbb R[x]$ a counterexample? (It lacks real roots but it isn't irreducible?) – Aryaman Maithani Jun 20 '20 at 21:23
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    Ah, good point. Not sure the answer can be salvaged, although the idea of thinking of the graph does have some value here – G Tony Jacobs Jun 20 '20 at 21:24
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    Maybe something like - if $f(x) = g(x)h(x)$, then the $y$-coordinate of any point on the graph of $y=f(x)$ is the product of the $y$-coordinates of the other two (possibly equal) $y$-coordinates on the same vertical line. This phenomenon will remain invariant under horizontal shifts. – Aryaman Maithani Jun 20 '20 at 21:29
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Very simple: for any $a$ in a ring $R$, the map $\;\begin {aligned}[t]R[X]&\longrightarrow R[X]\\ f(X)&\longmapsto f(X+a)\end{aligned}$ is a ring isomorphism ( the inverse isomotrphism maps $f(X)$ onto $f(X-a)$).

Therefore it maps irreducible polynomials onto irreducible polynomials and vice-versa.

Bernard
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