Finding integral domains.

Question

For a ring $R$ and an element $a\in R$, we denote the ideal generated by $a$ as $<a>$. With this notation we need to find which of the rings are integral domains ?

a.) $\Bbb Z[i]/(2)$, $\Bbb Z$ is the ring of integers.

b.) $\Bbb Q[X]/(X^4-5X+4)$, $\Bbb Q$ is the ring of integers.

c.) $\Bbb Z_{5}[X]/(X^2+X+1)$

d.) $\Bbb Z[X]/(3)$

Now to solve these kind of questions i use the result that if $R$ is commutative ring with unity, and $f(x)$ is irreducible polynomial on $R$, then $R[x]/(f(x))$ is isomorphic to $R$[roots of polynomial $f(x)$ over $C$]. But this does not help much in this case. Can you please tell how to approach this type of problem. My main point of asking this question is that how these sets a,b,c,d will look? And which result to use to verify or discard them as integral domains.

Answer 1

If $P\subset R$ is a prime ideal, then $R/P$ is an integral domain. If $P=(f)$, for some prime element (irreducible = prime, if $R$ is a PID) $f$, then $P$ is a prime ideal.

A nice trick is to try to write a quotient as a quotient of some polynomial, for instance:

• $\Bbb Z[i]\cong\Bbb Z[x]/(x^2+1)$, hence $$\Bbb Z[i]/(2)\cong \Bbb Z[x]/(2,x^2+1)$$ $x^2+1$ is irreducible in $\Bbb Z[x]$, but $x^2-1=x^2+1-2\in (2,x^2+1)$, so that $(2,x^2+1)=(2,x^2-1)$. Therefore $$\Bbb Z[i]/(2)\cong\Bbb Z[x]/(2,x^2-1)\cong\Bbb Z_2[x]/(x^2-1)$$ $x^2-1$ is reducible, so this ring has non trivial zero-divisors. This corresponds to the fact that $(1-i)(1+i)=2$

• Try to either prove that $X^4-5X+4$ is irreducible, or find a root and factor it. If it can be factored you'll know that the ring has non trivial zero-divisors.

• $\Bbb Z_5$ is a field, so $\Bbb Z_5[X]$ is a PID. Either show that $X^2+X+1$ is irreducible, and conclude that you have an integral domain (in fact a field!), or find a root and conclude that the ring is not an integral domain.

• Same as the point above, since $\Bbb Z_3$ is a field.