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

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.

Answer 1

This argument looks very good, on the whole.

Just a few small points.

• It's not necessarily true that each $p(x)$ is mapped to either $(a,b,c)$ or $(a+1,b+1,c+1)$, even for $p(x)$ non-constant. Let $p(x)=x+x^2$, and for any choice of $f$:

$$f\big(p(x)\big)=f(x)+f(x^2)=f(x)+f(x)=2f(x)=0.$$

• You claim that $(a+1,b,c)$ is not in the image. But what if $(a,b,c)=(1,0,0)$? Then $(a+1,b,c)=(0,0,0)$. Of course, you can easily get around this. Perhaps the easiest way is to count the maximal number of elements in the image $f(\mathbb Z_2[x])$, which you've essentially done already. Then note that $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$ contains strictly more elements that $f(\mathbb Z_2[x])$.

Answer 2

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 $|k| < n$.