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Eisenstein's Criterion 📂Abstract Algebra

Eisenstein's Criterion

Theorem 1

f(x)=anxn++a0Z[x]f(x) = a_{n} x^{n} + \cdots + a_{0 } \in \mathbb{Z} [ x ], if it satisfies the following conditions for the primes pZp \in \mathbb{Z} and k=0,1,2,,n1k = 0,1,2, \cdots , n-1, then f(x)f(x) is irreducible over Q\mathbb{Q}.

  • (i): an≢0(modp)a_{n} \not\equiv 0 \pmod{p}
  • (ii): ak0(modp)a_{k} \equiv 0 \pmod{p}
  • (iii): a0≢0(modp2)a_{0} \not\equiv 0 \pmod{p^2}

Description

It has significance as a very simple judgment method for integer polynomials of the form f(x)=axn+bf(x) = ax^{n} + b. It can be usefully discussed in relation to algebraic numbers as a judgment method for Q\mathbb{Q}.

Example

Show that f(x)=25x59x43x212f(x) = 25 x^{5} - 9 x^4 - 3 x^2 - 12 is irreducible over Q\mathbb{Q}.

Solution

  • (i) 25≢0(mod3)25 \not\equiv 0 \pmod{3}
  • (ii) 930(mod3)-9 \equiv -3 \equiv 0 \pmod{3}
  • (iii) 12≢0(mod9)12 \not\equiv 0 \pmod{9}

Applying Eisenstein’s criterion to p=3p=3 shows that f(x)f(x) is irreducible.

Proof

If f(x)Z[x]f(x) \in \mathbb{Z} [ x ] is said to be f(x)=R(x)S(x)f(x) = R(x) S(x) for R(x)R(x) of degree r<nr<n and S(x)S(x) of degree s<ns<n, then R(x),S(x)Q[x]    R(x),S(x)Z[x] R(x) , S(x) \in \mathbb{Q} [ x ] \iff R(x) , S(x) \in \mathbb{Z} [ x ] is known. Assuming from Z[x]Z[x] that f(x)=(brxr++b0)(csxs++c0) f(x) = (b_{r} x^{r} + \cdots + b_{0}) (c_{s} x^{s} + \cdots + c_{0}) satisfies the three conditions (i), (ii), and (iii), condition (iii) means that b0c0=a0≢0(modp2) b_{0} c_{0 } = a_{0} \not\equiv 0 \pmod{p^2} so b0b_{0} and c0c_{0} are not both b0c0p0(modp) b_{0} \equiv c_{0} \equiv p \equiv 0 \pmod{p} Instead, considering the case where one is congruent, {b0≢0(modp)c00(modp)\begin{cases} b_{0} \not\equiv 0 \pmod{p} \\ c_{0} \equiv 0 \pmod{p} \end{cases} then from condition (i), brcs=an≢0(modp) b_{r} c_{s} = a_{n} \not\equiv 0 \pmod{p} must be {br≢0(modp)()cs≢0(modp) \begin{cases} b_{r} \not\equiv 0 \pmod{p} \qquad \cdots (\star) \\ c_{s} \not\equiv 0 \pmod{p} \end{cases}

Now, if we call the smallest value among kk that satisfies ck≢0(modp)c_{k} \not\equiv 0 \pmod{p} as mm, then am=b0cm+b1cm1++{bmc0,rmbrcmr,r<m a_{m} = b_{0 } c_{m} + b_{1} c_{m-1} + \cdots + \begin{cases} b_{m} c_{0} & , r \ge m \\ b_{r} c_{m-r} & , r<m \end{cases} From ()(\star), b0≢0(modp)b_{0} \not\equiv 0 \pmod{p} and by the definition of mm, cm≢0(modp)c_{m} \not\equiv 0 \pmod{p}, therefore, cm1c00(modp) c_{m-1} \equiv \cdots \equiv c_{0} \equiv 0 \pmod{p} Thus, am≢0(modp) a_{m} \not\equiv 0 \pmod{p} and according to conditions (i) and (ii), m=nm=n must be.

Eventually, it is sm=ns \ge m = n, which contradicts the premise that s<ns < n.


  1. Fraleigh. (2003). A first course in abstract algebra(7th Edition): p215. ↩︎