Gauss Lemma Proof

There are two related results that are commonly called “Gauss Lemma”. The first is that the product of primitive polynomial is still primitive. The second result is that a primitive polynomial is irreducible over a UFD (Unique Factorization Domain) D, if and only if it is irreducible over its quotient field.

Gauss Lemma: Product of primitive polynomials is primitive

If D is a unique factorization domain and f,g\in D[x], then C(fg)=C(f)C(g). In particular, the product of primitive polynomials is primitive.


(Hungerford pg 163)

Write f=C(f)f_1 and g=C(g)g_1 with f_1, g_1 primitive. Consequently \displaystyle C(fg)=C(C(f)f_1C(g)g_1)\sim C(f)C(g)C(f_1g_1).

Hence it suffices to prove that f_1g_1 is primitive, that is, C(f_1g_1) is a unit. If f_1=\sum_{i=0}^n a_ix^i and g_1=\sum_{j=0}^m b_jx^j, then f_1g_1=\sum_{k=0}^{m+n}c_kx^k with c_k=\sum_{i+j=k}a_ib_j.

If f_1g_1 is not primitive, then there exists an irreducible element p in D such that p\mid c_k for all k. Since C(f_1) is a unit p\nmid C(f_1), hence there is a least integer s such that \displaystyle p\mid a_i\ \text{for}\ i<s\ \text{and}\ p\nmid a_s.

Similarly there is a least integer t such that \displaystyle p\mid b_j\ \text{for}\ j<t\ \text{and}\ p\nmid b_t.

Since p divides \displaystyle c_{s+t}=a_0b_{s+t}+\dots+a_{s-1}b_{t+1}+a_sb_t+a_{s+1}b_{t-1}+\dots+a_{s-t}b_0, p must divide a_sb_t. Since every irreducible element in D (UFD) is prime, p\mid a_s or p\mid b_t. This is a contradiction. Therefore f_1g_1 is primitive.

Primitive polynomials are associates in D[x] iff they are associates in F[x]

Let D be a unique factorization domain with quotient field F and let f and g be primitive polynomials in D[x]. Then f and g are associates in D[x] if and only if they are associates in F[x].


(\impliedby) If f and g are associates in the integral domain F[x], then f=gu for some unit u\in F[x]. Since the units in F[x] are nonzero constants, so u\in F, hence u=b/c with b,c\in D and c\neq 0. Thus cf=bg.

Since C(f) and C(g) are units in D, \displaystyle c\sim cC(f)\sim C(cf)=C(bg)\sim bC(g)\sim b.

Therefore, b=cv for some unit v\in D and cf=bg=vcg. Consequently, f=vg (since c\neq 0), hence f and g are associates in D[x].

(\implies) Clear, since if f=gu for some u\in D[x]\subseteq F[x], then f and g are associates in F[x].

Primitive f is irreducible in D[x] iff f is irreducible in F[x]

Let D be a UFD with quotient field F and f a primitive polynomial of positive degree in D[x]. Then f is irreducible in D[x] if and only if f is irreducible in F[x].


(\implies) Suppose f is irreducible in D[x] and f=gh with g,h\in F[x] and \deg g\geq 1, \deg h\geq 1. Then g=\sum_{i=0}^n(a_i/b_i)x^i and h=\sum_{j=0}^m(c_j/d_j)x^j with a_i, b_i, c_j, d_j\in D and b_i\neq 0, d_j\neq 0.

Let b=b_0b_1\dots b_n and for each i let \displaystyle b_i^*=b_0b_1\dots b_{i-1}b_{i+1}\dots b_n. If g_1=\sum_{i=0}^n a_ib_i^* x^i\in D[x] (clear denominators of g by multiplying by product of denominators), then g_1=ag_2 with a=C(g_1), g_2\in D[x] and g_2 primitive.

Verify that g=(1_D/b)g_1=(a/b)g_2 and \deg g=\deg g_2. Similarly h=(c/d)h_2 with c,d\in D, h_2\in D[x], h_2 primitive and \deg h=\deg h_2. Consequently, f=gh=(a/b)(c/d)g_2h_2, hence bdf=acg_2h_2. Since f is primitive by hypothesis and g_2h_2 is primitive by Gauss Lemma, \displaystyle bd\sim bdC(f)\sim C(bdf)=C(acg_2h_2)\sim acC(g_2h_2)\sim ac.

This means bd and ac are associates in D. Thus ubd=ac for some unit u\in D. So f=ug_2h_2, hence f and g_2h_2 are associates in D[x]. Consequently f is reducible in D[x] (since f=ug_2h_2), which is a contradiction. Therefore, f is irreducible in F[x].

(\impliedby) Conversely if f is irreducible in F[x] and f=gh with g,h\in D[x], then one of g, h (say g) is a unit in F[x] and thus a (nonzero) constant. Thus C(f)=gC(h). Since f is primitive, g must be a unit in D and hence in D[x]. Thus f is irreducible in D[x].


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