Quotient ring of complex polynomials and ideal domain

Question

Let f(X) = X^2 − 2X + 5 ∈ C[X] and the ideal generated by f(X) be I = f(X)C[X]. (where C(X) is the set of complex polynomials) Prove that the quotient ring C[X]/I is not an integral domain.

Since C(X) is commutative and has multiplicative identity 1, so does the quotient ring, satisfying (ID1).

I need to disprove: (ID2): whenever a, b ∈ C[X]/I satisfy ab=0, either a=0 or b=0 or both, or (ID2'): whenever a, b ∈ C[X]/I are non-zero, ab is also non zero

And that's where I get stuck, can anyone help me?

Answer 1

Hint. Remember that in the ring modulo $f(X)$, you can take $f(X)$ to be the zero element. So, given that complex numbers are allowed, can you factorise $f(X)$ into two polynomials?

(But not $f(X)=(1)f(X)$ or similar, as then one of the factors is zero.)

Answer 2

$\mathbb{C}[X]$ is factorial, so the question of whether $f$ is prime is the same as whether $f$ is irreducible.

So you need to show that $f$ isn't irreducible.

Answer 3

Hint $\ \Bbb Z/6\,$ is not a domain since $\,2\cdot 3 = 0,\ 2\neq 0,\ 3 \neq 0.\ $ Exactly the same thing happens in your example because over $\,\Bbb C\,$ every quadratic is reducible $\, f = gh,\,$ therefore $\,g,h\,$ will be nontrivial zero-divisors in $\,\Bbb C[x]/f,\,$ just as were $\,2,3\,$ above.