Determine which of the following rings are fields.

Question

Have I done it correctly?

Determine which of the following rings are fields:

a) $(\mathbb{Z}/2\mathbb{Z})[x]$/$\large_{(x^2+1)}$

b)$(\mathbb{Z}/3\mathbb{Z})[x]$/$\large_{(x^2+1)}$

My resolution:

a) We know that $(\mathbb{Z}/2\mathbb{Z})$ is a field. Then, $(\mathbb{Z}/2\mathbb{Z})[x]$ is a PID. We know that $(x^2+1)$ is maximal if the polynomial $x^2+1$ is irreducible. We have to search for the roots of the form $[a]\in (\mathbb{Z}/2\mathbb{Z})$, such that $[{p(a)]}=[a^2+1]=[0]$. The elements of $(\mathbb{Z}/2\mathbb{Z})$ are $[0]$ and $[1]$:$$ [p(0)]=[0+1]=[1]\neq[0], $$$$ [p(1)]=[1+1]=[2]=[0] $$

so $x^2+1$ is reducible.

b) We know that $(\mathbb{Z}/3\mathbb{Z})$ is a field. Then, $(\mathbb{Z}/3\mathbb{Z})[x]$ is a PID. We know that $(x^2+1)$ is maximal if the polynomial $x^2+1$ is irreducible. We have to search for the roots of the form $[a]\in (\mathbb{Z}/3\mathbb{Z})$, such that $[{p(a)]}=[a^2+1]=[0]$. The elements of $(\mathbb{Z}/3\mathbb{Z})$ are $[0]$, $[1]$ and $[2]$:$$ [p(0)]=[0+1]=[1]\neq[0], $$$$ [p(1)]=[1+1]=[2]\neq[0] $$ $$ [p(2)]=[4+1]=[5]=[2]\neq[0] $$

so $x^2+1$ is irreducible $\implies (\mathbb{Z}/3\mathbb{Z})[x]$/$\large_{(x^2+1)}$ is a field.

Answer 1

Yes! $F[X]/(f(x))$ ($F$ field ) is a field if and only if $f(x)$ is irreducible in $\mathbf{F}[X]$. Then what you've done is correct!

Answer 2

More concisely, for part a), observe that $\mathbb{Z}/2\mathbb{Z}[x]/(x^2+1)$ is not an integral domain, because $(\overline{x+1})(\overline{x+1})=\overline{x^2+2x+1}=\overline{x^2+1}=0$ in the quotient ring. And $\overline{x+1}$ is non-zero element. If it is not an integral domain, it surely cannot be a field.