Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.10
Solution:
(1) We claim that this subset $S$ is an ideal. To that end, suppose $\alpha = 3a + xp(x)$, $\beta = 3b + xq(x) \in S$ and $\tau = r + xt(x) \in \mathbb{Z}[x]$. Note that $$\alpha – \beta = 3(a-b) + x(p(x) – q(x)) \in S$$ and that $0 \in S$, so that $S$ is a subgroup. Now $$\alpha\beta = 9ab + x(3bp(x) + 3aq(x) + xp(x)q(x)) \in S,$$ so that $S$ is a subring. Finally, $$\alpha\tau = 3ar + x(3at(x) + rp(x) + xp(x)t(x)) \in S.$$ Thus $S$ is an ideal.
(2) Note that $x$ is in this set, but $xx = x^2$ is not. Since this subset is not closed under multiplication, it is not a subring, much less an ideal.
(3) We claim that this set $S$ is an ideal. To that end, let $\alpha = x^3p(x)$, $\beta = x^3q(x) \in S$ and $t(x) \in \mathbb{Z}[x]$. Note that $0 \in S$, and $$\alpha – \beta = x^3(p(x) – q(x)) \in S.$$ Thus $S$ is a subgroup. Moreover, $$\alpha\beta = x^6 p(x)q(x) \in S,$$ so that $S$ is a subring. Finally, $$\alpha t(x) = x^3 p(x)t(x) \in S,$$ so that $S$ absorbs $\mathbb{Z}[x]$. Thus $S$ is an ideal.
(4) Note that $x^2$ is in this set, but $xx^2 = x^3$ is not. Since this subset does not absorb $\mathbb{Z}[x]$ on the right, it is not an ideal.
(5) We claim that this set is an ideal. To that end, note that a polynomial $p(x)$ is in this set $S$ precisely when $p(1) = 0$. Suppose $p,q \in S$ and $t \in \mathbb{Z}[x]$. Clearly $0 \in S$ and $$(p-q)(1) = p(1) – q(1) = 0,$$ so that $p-q \in S$; thus $S$ is a subring. Moreover, $$(pt)(1) = p(1)t(1) = 0,$$ so that $S$ absorbs $\mathbb{Z}[x]$. Thus $S$ is an ideal.
(6) Let $p(x) = x^2 – 1$ and $q(x) = x$. Note that $p^\prime(x) = 2x$, so that $p$ is in this subset. However, $$(pq)^\prime = p^\prime q + q^\prime p,$$ so that $$(pq)^\prime(0) = p^\prime(0)q(0) + q^\prime(0)p(0) = p(0) = -1.$$ Since this subset does not absorb $\mathbb{Z}[x]$ on the right, it is not an ideal.
