# Show that the polynomial $X^5 + X^3 + \bar{1}$ in $(\mathbb{Z}/2\mathbb{Z}[X])$ is irreducible

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?

• Your answer is not correct, since a polynomial of degree $5$ over a field can be reducible although it has no roots. May 5 '19 at 15:54
• @DietrichBurde forgot its only for degree of $2$ or $3$.. May 5 '19 at 15:55
• What do you mean by $(\mathbb{Z}/2\mathbb{Z}[X])$ ? (how can you quotient $\mathbb{Z}$ by $2\mathbb{Z}[X] ) ?$ May 5 '19 at 15:56
• A typo for $\Bbb F_2[X]$ with $\Bbb F_2=\Bbb Z/2$. May 5 '19 at 15:57
• In other words, $(\mathbb{Z}/2\mathbb{Z})[X]$ May 5 '19 at 15:58

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).