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Is $x^2$ irreducible in $\mathbb{Z}/2\mathbb{Z}[x]$? I claim it is. $0\cdot 0 = 0, 1 \cdot 1 = 1 \implies x^2 \equiv x$, so the factorization $x^2 = x\cdot x$ is not a factorization into proper divisors. Saying that $x^2$ is reducible would be the same thing as saying that $1$ is reducible because $1 = 1\cdot 1$.

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    $\begingroup$ This is a bit of a subtle mistake: you are confusing polynomials as formal objects with their associated functions. That is, $x^{2}$ and $x$ are equal as polynomial functions on $\mathbb{Z}/2\mathbb{Z}$, but not as elements of $\mathbb{Z}/2\mathbb{Z}[x]$. $\endgroup$ Dec 15 '16 at 1:15
  • $\begingroup$ So it is not the case that $x^2 \equiv x$, merely $x^2 = x\ \forall x$? $\endgroup$
    – Jamie
    Dec 15 '16 at 1:34
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    $\begingroup$ The definition of irreducible is that $a=bc$ implies that either $b$ or $c$ is a unit in that ring, $x$ is not a unit in $\mathbb{F}_2[x]$ $\endgroup$
    – J. Doe
    Dec 15 '16 at 1:40
  • $\begingroup$ Artin's text says: $a$ is irreducible if it is not a unit, and its only divisors are units and associates. The only divisors of $x^2$ are 1 and $x$: $1$ is a unit, and $x$ is an associate of $x^2$ because $x^2=x$. $\endgroup$
    – Jamie
    Dec 15 '16 at 2:12
  • $\begingroup$ The distinction between polynomial functions and polynomials is very important - you should make sure you're comfortable with it now. Later on it becomes crucial in defining the general finite fields - you just can't get proper field extensions of $\mathbb F_p$ if your polynomials are only thought of as functions. $\endgroup$ Dec 15 '16 at 3:21

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