how to show that I is an ideal of Z[x]

Question

If we let I to be a set of all polynomials in Z[x] whose constant term is a multiple of 5. How can i proceed in proving that I is an ideal of Z[x] and what size is Z[x]/I ? I am still learning about ideals and its proofs, but my understanding is shallow. I couldn't reach to the required proof. Any help in understanding and proofing this statement?

Answer 1

Let $p \in I$, then it is of the form $p=a_0 + \sum_i a_i x^{i}$ where $a_i \in \mathbb{Z},$ and $ 5|a_0$. So, for any $p,q \in I, p+q = a_0 + b_0 + \sum_i (a_i + b_i) x^{i}.$ Here we have $5 | (a_0 +b_0)$ clearly. Hence, $p+q \in I$ as desired. Also, for $r \in \mathbb{Z}, r.p $ has a constant term $5 |r a_0, $ so $\forall r \in \mathbb{Z}, r.p\in I.$ Thus $I$ is an ideal.

Answer 2

In order to show $$I =\{f \in \mathbb{Z}[x] \mid \text{5 divides f(0)}\} = \{xf + 5a \mid f \in \mathbb{Z}[x], a \in \mathbb{Z}\}$$ is an ideal, we need to show that $I$ satisfies the ideal axioms:

1. $\forall a, b \in I, a - b \in I$.
2. $\forall a \in \mathbb{Z}[x], r \in I, ra \in I$.

For the first one choose $f = xf^{\prime} + 5a, g = xg^{\prime} + 5b \in I$. $$ f - g = (xf^{\prime} + 5a) - (xg^{\prime} + 5b) = x(f^{\prime} - g^{\prime}) + 5(a - b) \in I.$$

For the absorptive property, choose $g = xg^{\prime} + b \in \mathbb{Z}[x], f = xf^{\prime} + 5a \in I $ $$ f \cdot g = (xf^{\prime} + 5a)(xg^{\prime} + b) = x^2 f^{\prime} g^{\prime} + bxf^{\prime} + 5a xg^{\prime} + 5(ab) \in I.$$ Thus $I$ satisfies the ideal axioms.