Is $\left\{ f\in F[x] | \deg(f) < m \right\} \cup \left\{0\right\}$ an ideal?

Question

Is $\left\{ f\in F[x] | \deg(f) < m \right\} \cup \left\{0\right\}$ an ideal?

1. $0$ is clearly in the set
2. Easy to see that $a,b\in I \implies a+b\in I$
3. The last demand is that $a\in I, r\in R \implies ar,ra \in I$. But, if I understand correctly I may choose $h\in F[x]$ with $\deg(h) > m$ and so, $ah \notin I$

Am I right?

Answer 1

Yes, take $f\in\mathbb F[x]$ with $\deg(f)=m-1$, so $f\in I$, but $$\deg(xf) = \deg(f)+1 = m,$$ so $xf\notin I$.

Answer 2

Yes, you are right.

In fact, if an ideal $I$ in a ring of polynomials contains something different from $0$, then the degree of the polynomials is never bounded, for if it is bounded, then you can pick some $f$ with maximum degree, but then $xf\in I$, a contradiction.