By Randall R. Holmes

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**Example text**

Let a be the greatest common divisor of the numerators of the coefficients of g(x). Then g(x) = (a/b)g (x) with g (x) a primitive polynomial over Z. Doing the same thing for the polynomial h(x), multiplying the leading fractions and reducing, we conclude that f (x) = (a/b)g (x)h (x) for some primitive polynomials g (x), h (x) ∈ Z[x] with deg g (x) = deg g(x) and deg h (x) = deg h(x), and some relatively prime integers a and b. We claim that b = ±1. Suppose otherwise. Then b is divisible by a prime number p.

Clearing denominators in the equation f (c/d) = 0 gives 0 = a0 dn + a1 cdn−1 + · · · + an−1 cn−1 d + an cn so that −a0 dn = c(a1 dn−1 + · · · + an−1 cn−2 d + an cn−1 ). 56 This shows that c divides −a0 dn . Since c and d are relatively prime, it follows that c | a0 . Solving the above equation for −an cn instead similarly reveals that d | an . • We claim that the polynomial f (x) = 2 + x − 4x2 + 3x3 is irreducible over Q. If it is not irreducible, then it has a linear factor and hence a zero in Q.

Then ϕ(r) = ϕ(r + I) = 0, so that r ∈ ker ϕ = I. Thus, r + I = I. 6). Therefore, ϕ : R/I → im ϕ is an isomorphism and R/ ker ϕ = R/I ∼ = im ϕ. 2 Quotient same thing as homomorphic image Let R be a ring. The following theorem says that the notions “quotient of R” and “homomorphic image of R” amount to the same thing. 1 Theorem. If R/I (I R) is a quotient of R, then R/I is a homomorphic image of R, namely, the image under the canonical epimorphism π : R → R/I. Conversely, the image im ϕ = ϕ(R) of R under a homomorphism ϕ : R → R is isomorphic to a quotient of R, namely R/ ker ϕ.