Determine if the following are ideals.

Question

Determine if the following are ideals:

1. $I=\{f(x)\in \mathbb{Z}[x] : f'(0) = 0\}\subseteq \mathbb{Z}[x]$
2. $I = \{f(x)\in\mathbb{Z}[x] : f(0)=f'(0) = 0\}\subseteq \mathbb{Z}[x]$
3. $I =\{A \in M_2(\mathbb{R}) : A^n = 0 \text{ for some }n\in\mathbb{N}\} \subseteq M_2(\mathbb{R})$. In $M_2(\mathbb{R}),$ I'm pretty sure $0 := \left[\begin{matrix}0&0\\0&0\end{matrix}\right].$
4. $I = \{f\in F(\mathbb{R}) : f(1)=f(2)=0\}\subseteq F(\mathbb{R}).$

I know how to prove something is an ideal using the ideal test, but I find that very inconvenient, especially if something is not an ideal (e.g. $1$.) Isn't there some more efficient way of doing this?

Also, I have no idea how to solve $3.$ I think it might not be an ideal, but I'm not sure how to prove it. I'm pretty sure I first need to prove it's a subring.

Answer 1

Find counterexamples if the ring is not an ideal.

$1$ is not an ideal. Take $f(x) = x^2+x+4$ and $g(x)=x\in\mathbb{Z}[x].$ Let $h(x)=f(x)g(x)$. Then $h(x) = x^3+x^2+4x\not\in I$ as $h'(0)=4\neq 0.$

$2$ is an ideal. Let $f(x),g(x)$ in $I.$ Then $f(0)=f'(0)=0$ and $g(0)=g'(0)=0.$ We have $f(0)-g(0) = 0-0=0$ and $f'(0)-g'(0)=0-0=0.$ Thus $f(x)-g(x)\in I.$ Let $h(x)\in \mathbb{Z}[x].$ Then $h(0)f(0)=f(0)h(0)=0$ and $$[h(0)f(0)]'=h'(0)f(0)+h(0)f'(0)\\ =f'(0)h(0)+f(0)h'(0)\\ =[f(0)h(0)]'\\ =h'(0)\cdot 0+h(0)\cdot 0\\ =0.$$ Thus, $h(x)f(x),f(x)h(x)\in I$ so this is indeed an ideal.

$3$ is not an ideal. Pick $$A=\left[\begin{matrix}0&0\\1&0\end{matrix}\right]\in M_2(\mathbb{R}).$$ Then $$A^2 = \left[\begin{matrix}0&0\\1&0\end{matrix}\right]\cdot \left[\begin{matrix}0&0\\1&0\end{matrix}\right]\\ =\left[\begin{matrix}0&0\\0&0\end{matrix}\right]$$

so $A\in I.$

Now choose $C=\left[\begin{matrix}0&1\\1&0\end{matrix}\right]\in M_2(\mathbb{R}).$ Then $$CA = \left[\begin{matrix}0&1\\1&0\end{matrix}\right]\cdot \left[\begin{matrix}0&0\\1&0\end{matrix}\right]\\ =\left[\begin{matrix}1&0\\0&0\end{matrix}\right]$$

and $(CA)^2 = \left[\begin{matrix}1&0\\0&0\end{matrix}\right] \left[\begin{matrix}1&0\\0&0\end{matrix}\right] =\left[\begin{matrix}1&0\\0&0\end{matrix}\right]=CA\neq 0.$ Thus, there does not exist a natural number $n$ such that $(CA)^n = 0$ so $CA\not\in I\Rightarrow I$ is not an ideal.

$4$ is an ideal. Let $f(x),g(x)\in I.$ We have that $f(1)-g(1)=0\in I$ and $f(2)-g(2)=0\in I.$ So $f(x)-g(x)\in I$. Let $h(x)\in F(\mathbb{R}).$ Then $h(1)f(1)=f(1)h(1)=0h(1)=0$ and $h(2)f(2)=f(2)h(2)=0h(2)=0$ and so $h(x)f(x),f(x)h(x)\in I.$ Thus, this is an ideal.

Answer 2

For (1), as you say, you can just find a counterexample. For example, If $I$ is an ideal, then $1\in I$, but this implies $x\in I$, which is not true since its derivative at $0$ is $1$.

For (2), we only need to check that $I$ is an abelian group, and (since $\Bbb Z[x]$ is commutative) that $rx\in I$ for any $r\in \Bbb Z[x]$ and $x\in I$. If $f(x)\in I$ and $g(x)\in \Bbb Z[x]$, then $f(x)g(x)\in I$ since $(fg)'(0)=0$ (insert proof here).

For (3), consider the matrix $M\in I$ where $$M=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}. $$

Then left multiplication by the matrix shown gives

$$NM=\begin{pmatrix}0 & 0\\ 1 & 0\end{pmatrix}\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}=\begin{pmatrix}0 & 0 \\ 0 & 1\end{pmatrix}.$$

You can thus conclude $I$ is not a (left) ideal.

Etc. I will leave (4) to you!