Reduction modulo p

Question

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:

$f(X) = X^4 + 5X^2 − 2X − 3 = X^4 + X^2 + 1 = (X^2 + X + 1)^2\pmod 2$ $f(X) = X^4 + 5X^2 − 2X − 3 =X^4 + 2X^2 + X = X(X^3 + 2X + 1)\pmod 3$
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 $2$ and $3$?
Thank you in advance.

Answer 1

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!

Answer 2

If a polynomial is irreducible modulo $p$, then it is irreducible over the integers, for a factorization over the integers would induce a factorization modulo $p$.

A polynomial of degree $2$ or $3$ is irreducible over a field $F$ if and only if the polynomial has no zeros in the field.

By "trying everything" we can verify that $x^2+x+1$ has no zeros modulo $2$, and that $x^3+2x+1$ has no zeros modulo $3$.

Answer 3

First, note that if a polynomial $f$ is reducible, then you could witness these same factors modulo any prime. For instance, $x^2 - 1$ is reducible, and we see this mod 2: $(x+1)^2$ and mod 3: $(x+1)(x+2)$. Of course, it is possible that an irreducible factor of $f$ were further reduced mod $p$, for instance $x^2+1$ will appear as $(x+1)^2$ mod $2$.

Thus, if $f$ has an irreducible factor of degree $n$ it means that modulo any prime $p$ there must be a subset of the reducible factors such that the degree sum up to $n$.

In this particular case if $f$ were reducible, it would have a factor of degree $2$ or $3$. (Linear factors are easily found using the rational root test.(°)) But no subset of the irreducible factors mod $2$ sums up to $3$; whereas no subset of the degrees of the irreducible factors mod $3$ sums op to $2$. (°°) So this is impossible.

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(°) As Mathmo123 writes in his answer: it is also clear from the factorization mod 2 that there could not be a linear factor.

(°°) See the answer by André Nicolas for why the factors of $f$ are indeed irreducible.

Answer 4

Hint $\ 4$ is the only partition of $4$ having common refinements $2+2$ and $1+3$