1.2 Irreducible polynomials
How do we define more exotic fields? Well a good way is to take quotients of polynomial rings.
Let \(R\) be an integral domain. A non-zero polynomial \(p(x)\) of degree at least \(1\) in \(R[x]\) is said to be irreducible if whenever \(p(x)=f(x)g(x)\) with \(f(x),g(x) \in R[x]\) then one of \(f(x)\) or \(g(x)\) is in \(R\). Note this is slightly different to Definition 1.1.14.
The polynomial \(x^2+1\) is irreducible in \(\mathbb {Z}[x]\).
The polynomial \(x^2-1\) is not irreducible in \(\mathbb {Z}[x]\) since it is \((x-1)(x+1)\) and neither is a unit.
Let \(F\) be a field. Check that the only units in \(F[x]\) are given by polynomials of degree \(0\), i.e., they are elements of \(F\).
True or False: If \(p(x)\) is reducible in \(F[x]\) then there exists \(\alpha \in F\) such that \(f(\alpha )=0\) (i.e it has a root in \(F\)).
Let \(F\) be a field.
\(F[x]\) is a Euclidean domain and thus a Principal Ideal Domain (PID).
In a PID, the ideal generated by a irreducible element is prime.
Every non-zero prime ideal in a PID is maximal.
Every PID is a UFD.
Let me just remind you why \(F[x]\) is a Euclidean domain. This is because if you have two polynomials \(f(x),g(x)\) then we can do polynomial long division to write, \(f(x)=q(x)g(x)+r(x)\) where \(\deg (g) \leq \deg (f)\) and \(\deg (r) {\lt} \deg (g)\) (this last bit having a strict inequality is what is important).
Ok great, so if we want to create fields then we just need to find some irreducible polynomials and then just need to quotient out by it. In other words:
Let \(F\) be a field, and \(p(x)\) an irreducible polynomial in \(F[x]\). Then
is a field.
(In case you’ve forgotten.) What does this field look like? well its elements can be thought of as \(f(x)+a(x) p(x)\) where \(f(x), a(x) \in F[x]\). Here \(p(x)\) is the zero of this field, so \(f(x)\) and \(f+a(x)p(x)\) represent the same element in this field for any \(a(x)\).
Ok, so how do we check if a polynomial is irreducible? Let look at the case we will most care about, which is \(\mathbb {Q}[x]\). Now, we have the following Lemmas which you may have seen in the Galois theory course or algebra course, so we wont prove them:
A polynomial is irreducible in \(\mathbb {Z}[x]\) if and only if it is irreducible in \(\mathbb {Q}[x]\).
Here is a slightly different version for monic polynomials
Let \(f \in \mathbb {Z}[x]\) be monic and assume that \(f=gh\) with \(g,h \in \mathbb {Q}[x]\) which are also monic. Then \(g,h \in \mathbb {Z}[x]\).
Convince yourself that both these lemmas deserve to be named similarly.
If \(f(x)\) is a polynomial in \(\mathbb {Z}[x]\), let \(\bar{f}(x)\) denote its image in \(\mathbb {F}_p[x]\). A polynomial \(f(x)\) in \(\mathbb {Z}[x]\) is irreducible, if we can find some prime number \(p\) such that \(f(x)\) and \(\bar{f}(x)\) have the same degree and \(\bar{f}(x)\) is irreducible in \(\mathbb {F}_p[x]\).
This is the same as proving that, if \(f(x)\) is reducible in \(\mathbb {Z}[x]\) then it is also reducible in \(\mathbb {F}_p[x]\), which is obvious.
Lastly, we have:
Let \(f(x) \in \mathbb {Z}[x]\) be a monic polynomial of degree \(n{\gt}0\). Assume that there is a prime \(p\) and an integer \(a\) such that
for \(g(x) \in \mathbb {Z}[x]\). If \(g(a) \not\equiv 0 \bmod p\) then \(f(x)\) is irreducible modulo \(p^2\) and in \(\mathbb {Z}[x]\).
Assume for contradiction that
where WLOG \(r(x),s(x)\) are monic. Now, if we instead look at this modulo \(p\), we have \((x-a)^n \equiv r(x)s(x) \bmod p\) and therefore, since we are in \(\mathbb {F}_p[x]\) which is a UFD, we must have \(r(x) \equiv (x-a)^i \bmod p\) and \(s(x) \equiv (x-a)^j \bmod p\) with \(i+j=n\) and \(i,j{\gt}0\). If we now evaluate at \(x=a\) we see that
as \(i,j {\gt}0\). But if we now go back to
we see that setting \(x=a\) gives
which contradicts \(g(a) \not\equiv 0 \bmod p\).
Now, since we are irreducible modulo \(p^2\) we are also irreducible in \(\mathbb {Z}[x]\), since the same argument as in Proposition 1.2.13 applies.
Let
be a polynomial in \(\mathbb {Z}[x]\) (with \(n \geq 1)\) and let \(p\) be a prime number. Suppose \(a_n\) is not divisible by \(p\), \(a_{n-1},\dots ,a_0\) are all divisible by \(p\) and \(a_0\) is not divisible by \(p^2\). Then \(f(x)\) is irreducible.
Since \(a_n \not\equiv 0 \bmod p\), it is invertible. Now let \(b\) be any integer such that \(b \equiv a_n^{-1} \bmod p\), then \(bf(x)=x^n+pg(x)\) for some \(g(x) \in \mathbb {Z}[x]\). Moreover, \(g(0) \not\equiv 0 \bmod p\) since \(a_0\) is not divisible by \(p^2\). Therefore, we satisfy Shönemann’s Irreducibility Criterion and therefore \(bf(x)\) is irreducible and thus so is \(f(x)\).
The polynomial \(3x^4+10x+5\) is irreducible in \(\mathbb {Z}[x]\) by Eisensteins criterion with \(p=5\).
Let \(p\) be a prime and let
\[ \Phi _p(x):=\frac{x^p-1}{x-1}=x^{p-1}+\cdots +x+1. \]Then this is irreducible. Lets prove this:
Note that we cant apply Eisenstein’s criterion right away, but the trick is to consider
\[ \Phi _p(x+1)=x^{p-1}+px^{p-2}+\cdots +\binom {p}{2}x+p. \]Now, by properties of binomial coefficients, we see that all coefficients other than the leading one are divisible by \(p\) and the last coefficient isn’t divisible by \(p^2\), so Eisensteins criterion applies.
Prove that \(f(x) \in \mathbb {Z}[x]\) is irreducible if and only if \(f(x+a)\) is irreducible for any \(a \in \mathbb {Z}\).
Now we have more examples of fields:
Note that \(x^2-2\) is irreducible in \(\mathbb {Z}[x]\), so
\[ \frac{\mathbb {Q}[x]}{(x^2-2)} \]is a field.
\(\mathbb {R}[x]/(x^2+1)\) is also a field as \(x^2+1\) is irreducible over \(\mathbb {R}\). This field is isomorphic to \(\mathbb {C}\), but you need to be careful, its not *equal* to \(\mathbb {C}\), only isomorphic to. We will talk about this more below.
Let \(p(x)=x^2+x+1\) in \(\mathbb {F}_2[x]\). Then \(p(x)\) is irreducible in \(\mathbb {F}_2[x]\) and thus
\[ \mathbb {F}_2[x]/(x^2+x+1) \]is also a field.
Check that \(p(x)=x^2+x+1\) is irreducible in \(\mathbb {F}_2[x]\).
Lastly, here is a fact about finite fields we will use several times later but we wont prove.
For each integer \(n\) and each prime number \(p\), there is a unique finite field of size \(p^n\). We denote it by \(\mathbb {F}_{p^n}\). Conversely, every finite field has size \(p^n\) for some prime \(p\) and \(n \in \mathbb {Z}_{{\gt} 0}\).