Is the ideal $I=(t^2+1)$ of $\mathbb{Z}[t]$ prime or maximal?

Question

I'm asked to study whether the ideal $I=(t^2+1)$ of $\mathbb{Z}[t]$ is prime or maximal.

---

I believe the following argument shows that it is prime: If $ab\in I$ for $a,b\in \mathbb{Z}[t]$, then $$ab=(t^2+1)c$$ for some $c\in \mathbb{Z}[t]$, but $t^2+1$ is irreducible and so must divide either $a$ or $b$. Is this proof correct?

Regarding maximality, I'm stuck and would appreciate some help.

Answer 1

For solving such questions, it is better if work on $\Bbb{Z}[t]/(t^2+1)$ directly.

We have $$\Bbb{Z}[t]/(t^2+1)=\{a+bt:a,b\in \Bbb{Z},t^2=-1\}$$ therefore the product is $$(a+bt)(c+dt)=(ac-bd)+(ad+bc)t .$$ Then it is eqsy to see that if $(a+bt)(c+dt)=0$, then $a+bt=0$ or $c+dt=0$, which implies $\Bbb{Z}[t]/(t^2+1)$ is an integral domain, i.e. $(t^2+1)$ is prime.

Also $$(a+bt)^{-1}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}t$$ so, the ideal is not maximal.

Answer 2

Your proof that the ideal $I=(t^2+1)$ of $\mathbb{Z}[t]$ is prime is a valid proof, assuming you have the fact that $\mathbb{Z}[t]$ is a UFD.

To show $I$ is not maximal, it suffices to find a proper ideal $J$ which strictly contains $I$.

Consider the ideal $J=(t^2-1,2)$.

Since $t^2+1=(t^2-1)+2$, it follows that $t^2+1\in J$, hence $I\subseteq J$.

But $2\not\in I$ since any nonzero multiple of $t^2+1$ has degree at least $2$.

Since $I\subseteq J$, and $2\in J$ but $2\not\in I$, we get that $J$ strictly contains $I$.

Next suppose $1\in J$.

Then there must exist $a,b\in\mathbb{Z}[t]$ such that the equation $$a(t){\,\cdot\,}(t^2-1)+b(t){\,\cdot\,}2=1$$ is an identity.

Plugging in the value $t=1$, we get $b(1){\,\cdot\,}2=1$, contradiction, since $b(1){\,\cdot\,}2$ is an even integer.

Hence $1 \not\in J$, so $J$ is a proper ideal.

Thus $J$ is a proper ideal which strictly contains $I$, hence $I$ is not maximal.