Ideals $(X^2+1)$ and $(X^2+1, 7)$ of polynomial ring $\mathbb{Z}[X]$

Question

How can I show that generated ideals $(X^2+1)$ and $(X^2+1, 7)$ of polynomial ring $\mathbb{Z}[X]$ are a prime ideal and a maximal ideal, respectively?

Answer 1

To show $(X^2+1)$ is a prime ideal, imagine you took two polynomials with integer coefficients, $P(X)$ and $Q(X)$, and $P(X)\cdot Q(X) = (X^2+1)R(X)$ for some other polynomial $R(X)$. The polynomial $X^2+1$ is degree $2$ and has no roots in $\mathbb Z$, so it is irreducible... Can you take it from here?

To show that $\mathbb Z[X]\big/(X^2+1,7)$ is a field (and hence that $(X^2+1,7)$ is maximal), first you can prove that as rings $$ \frac{\mathbb Z[X]}{(X^2+1,7)}\cong \frac{\big(\mathbb Z\big/(7)\big)[X]}{(X^2+1)}, $$ and that $F=\mathbb Z/(7)$ is a field. As $-1$ is not a square in $F$, $F[X]\big/(X^2+1)\cong F(i)$ is a field extension of order $2$, but in particular $F(i)$ is a field, so $(X^2+1,7)$ is maximal.

Answer 2

Hint

Are you able to check that $$\Bbb{Z}[x]/\langle x^2+1, 7\rangle =\{ax+b\, | \, a,b \in \Bbb{Z}_7, \, x^2+1 \equiv 0\}?$$ This is a finite set with $7^2=49$ elements. If you want to check this is a field, you need to check the properties that define a field. One such property is to see if this has any zero divisor (finite integral domain is a field). \begin{align*} (ax+b)(cx+d) \equiv 0 \\ acx^2+(ad+bc)x+bd \equiv 0 \\ (ad+bc)x+(bd-ac) \equiv 0 && (\because x^2 \equiv -1) \end{align*} This implies \begin{align*} ad+bc & \equiv 0 \pmod{7}\\ bd-ac& \equiv 0 \pmod{7}. \end{align*}

Squaring and adding gives us \begin{align*} (ad+bc)^2+(bd-ac)^2 & \equiv 0 \pmod{7}\\ (ad)^2+(bc)^2+(bd)^2+(ac)^2 & \equiv 0 \pmod{7}\\ (a^2+b^2)(c^2 + d^2)& \equiv 0 \pmod{7} \end{align*}

See if you can check that at least one of $ax+b$ or $cx+d$ has to be the zero element?

Similar approach can be helpful for the first part if you can show that

$$\Bbb{Z}[x]/\langle x^2+1\rangle =\{ax+b\, | \, a,b \in \Bbb{Z}, \, x^2+1 \equiv 0\}$$