Factorization of fields to finitely generated ideals

Question

Let $K$ be a field and $I=\langle 2, x^2, x^3\rangle$ an ideal of $K[x]$. How does $K[x]/I$ look like? I think I have a quite poor grasp of how factorization works in rings and I would very much appreciate an explanation.

Answer 1

First, note that if $x^2 \in I$, then $x^3 = x \cdot x^2\in I$ by the definition of an ideal. So we can remove $x^3$ from the generators: $I = \langle 2, x^2 \rangle$.

Next, consider two cases:

• $2$ is a unit (invertible) in $K$. Then, since $2 \in I$, we have $2 \cdot {1 \over 2} = 1\in I$. An ideal containing $1$ coincides with the whole ring, since $\forall x \in K[x]: x = 1 \cdot x \in I$. In this case, $I = K[x]$ and $K[x]/I = 0$ (the zero ring).

• $2$ is not a unit in $K$, which can happen only if $2 = 0$ in $K$ (that is, the characteristic of the field is $2$). In this case we can remove $2$ from the set of generators, since any ideal contains $0$ by definition: $I = \langle x^2 \rangle$. Now, any element of $K[x]/I$ can be represented by $a + bx$, where $a,b \in K$ and the multiplication rule is derived from $x^2 \in I$, which translates to $x^2 = 0$ in the quotient ring: $(a+bx)(c+dx)=ac+(ad+bc)x+bdx^2 = ac+(ad+bc)x$. This formula, with the obvious addition rules, fully describes the ring $K[x]/I$.