Showing that a given subset is an ideal of a polynomial ring

Question

Let $\Bbb Z[x]$ be the set of polynomials in indeterminate $x$ with integer coefficients. Consider the subset $I = \{ 2f(x) + xg(x) | f(x), g(x) ∈ \Bbb Z[x] \}$. Show that $I$ is an ideal in $\Bbb Z[x]$.

I'm having a bit of trouble getting this started.

Answer 1

You will need to check that I satisfies the definition of an ideal. This means I needs to be a subgroup of the ring $\mathbb{Z}$[x], which is closed under left and right multiplication by elements $p(x)\in \mathbb{Z}$[x].

To show that I is indeed a subgroup it is enough to show it is closed under multiplication and subtraction. Suppose $p(x),q(x)\in I$ then: $$p(x)=2f_1(x)+xg_1(x)\tag{1}$$ $$q(x)=2f_2(x)+xg_2(x)\tag{2}$$ with $f_i,g_j\in \mathbb{Z}$[x]. Hence, $$p(x)-q(x)=2(f_1(x)-f_2(x))+x(g_1(x)-g_2(x))\tag{3}$$Since $\mathbb{Z}$[x] is closed under subtraction the difference is in I. You also need to show I is closed under multiplication f,but this is not bad either. Since $q(x)\in I\subset \mathbb{Z[x]}$ we have $$p(x)q(x)=2f_1(x)q(x)+xg_1(x)q(x)\tag{4}$$ where again this has the desired form since $\mathbb{Z[x]}$ is a ring($f_1(x)q(x)\in \mathbb{Z[x]}$ etc..) . So I is a subgroup of the ring. Notice we've only shown I is closed under multiplication between elements in I. To show that it is an ideal we also have to show that for $f(x)\in\mathbb{Z[x]}$ and $p(x)\in I$ that $p(x)f(x)\in I$ and $f(x)p(x)\in I$. I will leave this last part to you. Hope this helps.

Answer 2

Someone recently commented in another thread that whenever we have an ideal of a ring, it can be profitable to determine which homomorphism has this ideal as its kernel. In this case, we can use this interpretation to show that the set $I$ in your question is an ideal.

Define \begin{align*} \varphi: \mathbb{Z}[x] &\to \mathbb{Z}/2\mathbb{Z}\\ f(x) &\mapsto \overline{f(0)} \end{align*} where the bar indicates the remainder mod $2$. One can show that $\varphi$ is a homomorphism (indeed, it is the composition of the evaluation homomorphism $f(x) \mapsto f(0)$ and the quotient map $\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$) and that its kernel is the set $I$ in your question. Since the kernel of a homomorphism is always an ideal, this shows that $I$ is an ideal.