order of quotient ring

abstract-algebra field-theory ring-theory

In $\mathbb{Z}\left [ x \right ]$, let $I = \left \{ f\left ( x \right ) \in \mathbb{Z}\left [ x \right ]: f\left ( 0 \right ) \text{ is an even integer} \right \}$

In fact, $I=\left \langle x,2 \right \rangle$

How do I show that the order of $\mathbb{Z}\left [ x \right ]/I$ is 2?


Hint: Consider the ring homomorphism $\mathbb Z[x] \to \mathbb Z/ 2\mathbb Z$ given by $f(x) \mapsto f(0) \bmod 2$.

Equivalently, prove that $f(x) \equiv (f(0) \bmod 2) \bmod I$.


The order of the quotient by an ideal is the number of cosets the ideal has in the full ring, or equivalently, the number of equivalence classes modulo that ideal. Thinking in terms of cosets, we have that $1+I$ is the set of all polynomials $f$ in $\mathbb{Z}[x]$ with $f(0)$ odd. Since $f(0)$ must be odd or even, there can be no more cosets.

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