The ring of polynomials over a field with no linear term is not a UFD

Let F be a field. Prove that the subset R \subseteq F[x] consisting of all polynomials whose linear coefficient is zero is a subring. Prove also that R is not a unique factorization domain.


To show that R is a subring, we need to see that it is closed under subtraction and multiplication. To that end, let \alpha = a_0  + x^2a(x) and \beta = b_0 + x^2 b(x) be in R. Then \alpha - \beta = (a_0 - b_0) + x^2(a(x) - b(x)) \in R and \alpha\beta = a_0b_0 + x^2(b_0a(x)x^2 + a_0b(x)x^2 + x^4a(x)b(x)) \in R. So R is a subring.

Next, we claim that x^2 and x^3 are irreducible in R. To see this, note that no element of R has degree 1. If x^2 = p(x)q(x) factors in R, then computing the degree of both sides we have \mathsf{deg}(p(x)) + \mathsf{deg}(q(x)) = 2. Since neither of these degrees is 1, one must be 0, so that without loss of generality, p(x) is constant and thus a unit. So x^2 is irreducible. Similarly, x^3 is irreducible because if two nonnegative integers sum to 3 then one of them must be 0 or 1. So x^2 and x^3 are irreducible in R.

Now x^6 = x^2x^2x^2 = x^3x^3, so that x^6 has at least two distinct irreducible factorizations in R. Thus R is not a unique factorization domain.

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Comments

  • kem  On April 25, 2012 at 9:43 pm

    I think the 4th line should be \alpha\beta= a_0b_0+x^2(a_0b(x) + b_0a(x) +x^2a(x)b(x))\in R

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