TITLE: Show that there is no surjective ring homomorphism from $\mathbb Z_2[x]$ to $\mathbb Z_2 \times \mathbb Z_2\times \mathbb Z_2$
QUESTION [8 upvotes]: I saw this question as a bonus from a past exam, and here's my solution for verification.
I argued like so. I said suppose there is such a surjective homomorphism $f$,
then $f(0)=(0,0,0)$, $f(1)= (1,1,1)$ by ring homomorphism axioms. Suppose now that $f(x)= (a,b,c)$, where $a,b,c$ are either $0$ or $1$. Then $f(x^2)= (a^2,b^2,c^2)= (a,b,c)$ and same thing for $f(x^n)$ for any $n\geq 1$. Now this implies that any $p(x)$ will be mapped to either $(a,b,c)$ or $(a+1,b+1,c+1)$, depending on if they have a constant term (1). This means that $(a+1,b,c)$ for example is not in the image of $f$. Done.
Is this a good argument? Thanks in advance.

REPLY [5 votes]: Morgan O has pointed out the minor errors in your solution.  Here is a more "top down" way to look at it:
If $k$ is a field, and $p(X)\in k[X]$ is nonzero, then $k[X]/p(X)$ is a product of field extensions of $k$, each obtained by adjoining a solution of some irreducible factor of $p$.  So $k[X]/p(x)\cong k\times k \times k$ exactly when $p(X)$ is a product of three distinct linear polynomials.
Since, in our case, there are only two linear polynomials, we are done.  In general, we cannot have $k[X]$ mapping surjectively onto $k^n$ when $k$ is a finite field with $|k| < n$.