Number of ring homomorphisms from $\mathbb{Z}[X]/(X^2-X)$ to $\mathbb{Z}/6\mathbb{Z}$

Question

I want to determine the number of ring homomorphisms from $\mathbb{Z}[X]/(X^2-X)$ to $\mathbb{Z}/6\mathbb{Z}$.

If we suppose that such a homomorphism is not the zero homomorphism, then $f(1)=1\bmod6$. How do I use this to find all the others?

Answer 1

Call $R= \Bbb{Z}[X]/(X^2-X)$.

It is enough to choose $f(X)$ properly. After doing this choice, the unique way to define $f$ is $$f(\sum_i a_i X^i) = \sum_i a_i f(X)^i$$

Now, since $X^2=X$ in $R$, necessarily $f(X)^2=f(X)$ in $\Bbb{Z}/6 \Bbb{Z}$. Equivalently, $f(X) \in \{ 0,1, 3,4\}$: this shows that there are exactly 4 ring homomorphisms (one for every choice of $f(X)$).

Answer 2

Hint: A homomorphism $\Bbb Z[X]/(X^2-X)\to A$ is nothing but a homomorphism $\Bbb Z[X]\to A$ where $X^2-X$ "happens" to map to $0\in A$. And a homomorphism $\Bbb Z[X]\to A$ is nothing but a homomorphism $\Bbb Z\to A$ together with a choice where $X$ should map to.

Answer 3

If $1\mapsto1$, it is a $\mathbf Z$-algebra homomorphism from $\mathbf Z[X]/(X^2-X)$ to $\mathbf Z/6\mathbf Z$. This is the same as a $\mathbf Z$-algebra homomorphism from $\mathbf Z[X]$ to $\mathbf Z/6\mathbf Z$, which vanishes for $X^2-X$.

Now a $\mathbf Z$-algebra homomorphism from $\mathbf Z[X]$ to any $\mathbf Z$-algebra $B$ is entirely determined once you've chosen the image of $X$. Hence the $\mathbf Z$-algebra homomorphism from $\mathbf Z[X]/(X^2-X)$ to $\mathbf Z/6\mathbf Z$ correspond bijectively to the roots of $X^2-X$ in $\mathbf Z/6\mathbf Z$.

It is easy to check there are $4$ roots: $\;\{0,1,3,4\}$ and correspondingly $4$ homomorphisms.

Answer 4

Hint The ring is generated by $[1]$ and $[X]$, so to determine a homomorphism $\phi$ it suffices to determine the image of $[X])$. Now, $[X]$ is idempotent, hence $\phi([X])$ is, too. The idempotent elements of $\Bbb Z / 6 \Bbb Z$ are $[0], [1], [3], [4]$.

Alternatively, one can use that $\Bbb Z [X] / (X^2 - X)$ is isomorphic to $\Bbb Z \oplus \Bbb Z$ via the map $$f + (X^2 - X) \mapsto (f(0), f(2)) ,$$ and hence instead count the ring homomorphisms $\Bbb Z \oplus \Bbb Z \mapsto \Bbb Z / 6 \Bbb Z$.