Prove that $U = t · \mathbb{R}[t]$ is a maximal ideal in $\mathbb{R}[t]$

Question

I was studying for an exam and chanced upon this question in my textbook. I was a bit confused as to how we would go about trying to solve it.

Any help would be appreciated! :)

Prove that $U = t · \mathbb{R}[t]$ is a maximal ideal in $\mathbb{R}[t]$.

Answer 1

We assume that by $R[t]$ you mean the ring of polynomials with real coefficients. But assuming that $R$ is a field is enough. Edit: OP has clarified that it is the reals.

That $U$ is an ideal is a straightforward verification of properties.

We show now that any ideal $J$ which is a proper extension of $U$ is the whole ring. Let $f(t)$ be an element of $J$ which is not in $U$. Then the constant term of $f(t)$ is non-zero. If follows that $f(t)=a+tg(t)$ for some non-zero $a$ and some polynomial $g(t)$.

Thus by subtraction $a\in J$ and therefore $1\in J$ and therefore $J$ is the whole ring.

Answer 2

Here is a more structural approach to supplement André Nicolas's excellent answer. Let $R$ be a field. Consider the ring homomorphism $\phi: R[t] \rightarrow R$ given by $\phi(f(t)) = f(0)$. Like all evaluation maps, it is straightforward to see that $\phi$ is indeed a homomorphism. It is also straightforward to see that $\phi$ is surjective; note, for instance, that $\phi(t+r) = 0 +r = r$ for any $r \in R$.

Futhermore, one can show that $\langle t \rangle$, the ideal generated by $t$, is the kernel of this map. It is clear that $\langle t \rangle \subset \ker \phi$, since any element of $\langle t \rangle$ may be written $tk(t)$ for some $k(t) \in R[t]$, whence $\phi(tk(t)) = 0k(0) = 0$. Since $R$ is a field, $R[t]$ is a Euclidean domain, and so we can apply polynomial division by $t$ to conclude that $\ker \phi \subset \langle t \rangle$ as follows. Suppose $k(t) \in \ker(\phi)$. Then we apply polynomial division by $t$ to find $k(t) = tg(t) + r(t)$, where $r(t)$ must be constant since it must have degree less than $1$. But then $k(0) = 0g(0) + r(0) = 0$, so $r(0) = 0$ and hence $r = 0$. Thus, $k(t) = tg(t)$, and so $\ker(\phi) \subset \langle t \rangle$.

Hence, we conclude by the first isomorphism theorem for rings that $R[t]/\langle t \rangle \cong R$. Since $R$ is a field, this implies that $\langle t \rangle$ is maximal.