8
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – rschwieb
    Aug 30, 2014 at 12:14

2 Answers 2

6
$\begingroup$

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 $f(\mathbb Z_2[x])$.
$\endgroup$
3
  • 1
    $\begingroup$ 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. $\endgroup$
    – user172377
    Aug 29, 2014 at 3:17
  • 5
    $\begingroup$ @joe I don't mind if you call me "sir," but "madam" would be more technically correct :-) And glad to be of help. $\endgroup$ Aug 29, 2014 at 3:31
  • 1
    $\begingroup$ Oops sorry Madame :) Thanks again $\endgroup$
    – user172377
    Aug 29, 2014 at 13:27
5
$\begingroup$

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 $|k| < n$.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – user172377
    Aug 29, 2014 at 3:15
  • $\begingroup$ @zcn Well, it might be the empty product of field extensions :) $\endgroup$ Aug 29, 2014 at 3:30

You must log in to answer this question.