# Show that there is no surjective ring homomorphism from $\mathbb Z_2[x]$ to $\mathbb Z_2 \times \mathbb Z_2\times \mathbb Z_2$

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.

• The long story short is that the second ring has three quotients isomorphic to the field of two elements, but the first ring only has two. Commented Aug 30, 2014 at 12:14

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 $f(\mathbb Z_2[x])$.
• Thank you sir, that's exactly what I was trying to do subconsciously, to say that the number of elements in z2xz2xz2=8 is more than the image of any f(Z2[x]). And so instead of just counting I tried to come up with a general formula like (a+1b,c) not being in the image but I realized that wouldn't work as you pointed out.
– user172377
Commented Aug 29, 2014 at 3:17
• @joe I don't mind if you call me "sir," but "madam" would be more technically correct :-) And glad to be of help. Commented Aug 29, 2014 at 3:31
• Oops sorry Madame :) Thanks again
– user172377
Commented Aug 29, 2014 at 13:27

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 $|k| < n$.

• Thanks for all those facts, I haven't really seen field extensions but nevertheless, I think I sort of get the gist of what you explained.
– user172377
Commented Aug 29, 2014 at 3:15
• @zcn Well, it might be the empty product of field extensions :) Commented Aug 29, 2014 at 3:30