Reduction modulo p I am going to begin the Tripos part III at Cambridge in October (after going to a different university for undergrad) and have been preparing by reading some part II lecture notes. Here is an extract from some Galois theory notes: '$f(X) = X^4 + 5X^2 − 2X − 3 = X^4 + X^2 + 1 = (X^2 + X + 1)^2$ (mod 2) $f(X) = X^4 + 5X^2 − 2X − 3 =X^4 + 2X^2 + X = X(X^3 + 2X + 1)$ (mod 3) So f is irreducible, since f = gh implies deg g = 1 or deg g = 2, which is impossible by reduction modulo 2 and 3, respectively.' Could someone please explain the last sentence? Why is deg g=1 or 2 impossible by reduction modulo 2/3? Thank you in advance.

If $f$ is reducible, then in particular $f$ is reducible modulo every prime $p$ – since $$f(x)=g(x)h(x) \implies f(x) \equiv g(x)h(x) \pmod p $$for every $p$.

So if $f=gh$ where deg $g$ = $1$, then modulo $2$ we would expect $f$ to have a degree $1$ factor. Since this is not the case, we cannot have deg $g = 1.$

Similarly, if deg $g = 2$, then we would expect $f$ to have factors with degrees summing to $2$ modulo $3$. Since this is not the case, we cannot have deg $g=2$ \- so $f$ is irreducible.

As a side point, if you haven't discovered it yet, this website has a large collection of notes based on the Cambridge Tripos that may be worth looking at!