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Show that the polynomial $X^5 + X^3 + \bar{1}$ in $(\mathbb{Z}/2\mathbb{Z})[X]$ is irreducible. (Hint: if it were reducible, it would either have a root or be of the form $g(X) \cdot h(x)$, where deg$g(X) = 2$ and deg$h(X) = 3$) Recall that $\bar{a}$ is shorthand for the coset $a + 2\mathbb{Z} \space $in$ \space (\mathbb{Z}/2\mathbb{Z})[X]$ (for any $a \in \mathbb{Z}$)

This question is worth $7$ marks out of a possible $75$, this is my solution..

$f(X) = X^5 + X^3 + \bar{1}$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[X]$ iff it has no roots in $\mathbb{Z}/2\mathbb{Z}$. Since $f(\bar{0}) = \bar{1}$ and $f(\bar{1}) = \bar{1}$, it follows that $f(x)$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[X]$.

Would this be a sufficient answer? am I missing something?

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    $\begingroup$ Your answer is not correct, since a polynomial of degree $5$ over a field can be reducible although it has no roots. $\endgroup$ – Dietrich Burde May 5 '19 at 15:54
  • $\begingroup$ @DietrichBurde forgot its only for degree of $2$ or $3$.. $\endgroup$ – Ben Jones May 5 '19 at 15:55
  • $\begingroup$ What do you mean by $(\mathbb{Z}/2\mathbb{Z}[X])$ ? (how can you quotient $\mathbb{Z} $ by $ 2\mathbb{Z}[X] ) ? $ $\endgroup$ – J. W. Tanner May 5 '19 at 15:56
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    $\begingroup$ A typo for $\Bbb F_2[X]$ with $\Bbb F_2=\Bbb Z/2$. $\endgroup$ – Dietrich Burde May 5 '19 at 15:57
  • $\begingroup$ In other words, $(\mathbb{Z}/2\mathbb{Z})[X]$ $\endgroup$ – J. W. Tanner May 5 '19 at 15:58
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Well, its sufficient to check that the polynomial cannot be divided by an irreducible polynomial of degree 1, $X$ and $X+1$ (this is clear since the polynomial has no zero in the prime field), and degree 2, $X^2+X+1$ (which is the only irreducible polynomial of this degree).

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