Sachchidanand Prasad

19-04-2023

Problem: Which of the following rings is an integral domain.

  1. $\frac{\mathbb{Z} [x]}{\left\langle x^2 + x + 1 \right\rangle }$.
  2. $\frac{\mathbb{Z} [x]}{\left\langle x^2 - 2 \right\rangle }$.
  3. $\frac{\mathbb{Z} [x]}{\left\langle x^2 + 5x + 6 \right\rangle }$.
  4. $\frac{\mathbb{Z}[x]}{\left\langle x^7 + 1 \right\rangle }$.

Solution: We recall the following theorem.

Let $R$ be a commutative ring with unity. Let $I$ be an ideal of $R$. Then $R/I$ is an integral domain if and only if $I$ is a prime ideal.
Here the ring is $\mathbb{Z} $ and we recall that for $p(x)\in \mathbb{Z} [x]$, the ideal $\left\langle p(x) \right\rangle $ is a prime ideal if and only if $p(x)$ is irreducible. So we need to check in each option whether the polynomial is irreducible or not.

In option $(A)$ the polynomial is $p(x) = x^2 + x + 1$, which is a quadratic polynomial and hence irreducibility is equivalent to have a root in $\mathbb{Z} $. Since $p(x)$ does not have a root in $\mathbb{Z} $ and hence it is irreducible. Therefore, $\mathbb{Z} [x]/\left\langle x^2 + x + 1 \right\rangle $ is an integral domain.

In option $(B)$ the situation is same as in option $(A)$ since the polynomial is quadratic. As the polynomial $x^2 - 2$ does not have a root in $\mathbb{Z} $, the ideal generated by it is a prime ideal and hence $\mathbb{Z} [x]/\left\langle x^2 - 2 \right\rangle $ is an integral domain.

In option $(C)$, the polynomial is $x^2 + 5x + 6 = (x+2)(x+3)$ which is reducible and hence the ideal is not a prime ideal and hence $\mathbb{Z} [x]/\left\langle x^2 + 5x + 6 \right\rangle $ is not an integral domain.

Finally, in option $(D)$ the polynomial $\left\langle x^7 + 1 \right\rangle $ is reducible as $-1$ is a root of $x^7 + 1$ and hence, the ideal generated by this polynomial is not a prime ideal and therefore, $\mathbb{Z} [x]/\left\langle x^7 +1 \right\rangle $ is not an integral domain.

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