Is this structure an ideal?

Question

If we view $\mathbb R$ as a subset of $\mathbb R[x]$ by just identifying $r$ with $r + 0x + 0x^2$, etc.

I know then that $\mathbb R$ is a subring of $\mathbb R[x]$. But is it an ideal?

Answer 1

No because it isn't closed under multiplication by elements of $\mathbb R[X]$. Just take $X\in\mathbb R[X]$ and $1\in\mathbb R$: if $\mathbb R$ would be an ideal of $\mathbb R[X]$ then $1\cdot X=X$ should stay in $\mathbb R$. Absurd!

Answer 2

No, it is not. Whatever polynomial no constant $p(x)\in R[x]$ implies that $a p(x)\not\in R$ where $a\in R$. Then $R$ is not an ideal.