0
$\begingroup$

Number of Ring homomorphism form

$$\displaystyle\mathbb{Z[x,y]}\;\;\;\;\; to\;\;\;\; \frac{\mathbb{F_2[x]}}{\langle x^3+x^2+x+1 \rangle}$$

As we can see that $\displaystyle \frac{\mathbb{F_2[x]}}{\langle x^3+x^2+x+1 \rangle}$ is not a field because the ideal $\langle x^3+x^2+x+1 \rangle $ is not a irreducible element in $\mathbb{F_2[x]}$.Does i have to calculate the ideals of $\mathbb{Z[x,y]}$?,i have no idea how to proccede please help.

Thankyou

$\endgroup$
  • $\begingroup$ A homomorphism is uniquely determined by where it sends a generating set of the domain. The domain is generated by $1, x, y$, and $\Bbb F_2[x]/(x^3+x^2+x+1)$ has a finite number of elements, so there are only a finite number of functions from $\{1, x, y\}$ to $\Bbb F_2[x]/(x^3+x^2+x+1)$ in the first place. And nowhere near all of them give rise to homomorphisms. What stops you from, you know, just checking all of them? If you want, you can use the Chinese remainder theorem to rewrite $\Bbb F_2[x]/(x^3+x^2+x+1)$ as a direct product of fields to make it even easier. $\endgroup$ – Arthur Nov 18 '19 at 9:31
  • $\begingroup$ what these functions will look like?. $\endgroup$ – TheStudent Nov 18 '19 at 9:34
  • $\begingroup$ For instance, sending $1$ to $x^2$, sending $x$ to $x+1$ and sending $y$ to $1$ would be one such function. It doesn't give rise to a homomorphism,in particular because $1$ is not sent to $1$ (or, if your definition of ring homomorphisms is a bit more slack, because $1$ isn't sent to an idempotent element). Another function sends $1$ to $1$, $x$ to $x^2$ and $y$ to $x^2+1$. This one will give rise to a homomorphism, but that must be proven. $\endgroup$ – Arthur Nov 18 '19 at 9:35
  • $\begingroup$ i am getting $(x^2+1)$ and $(x+1)$ as factors of $(x^3+x^2+x+1)$ what is next how can i apply CRT here,because both are reducible over $\mathbb{F_2}$ $\endgroup$ – TheStudent Nov 18 '19 at 9:41
  • $\begingroup$ You're right, the CRT will possibly be a bit tricky here. To apply it, you want to factor $x^3+x^2+x+1$ into two (or more) coprime factors. That looks like it could be difficult here. $\endgroup$ – Arthur Nov 18 '19 at 9:43
3
$\begingroup$

Adding to the comment thread above, here is a justification for why the answer is $2^6$:

There are $8$ elements in $\Bbb F_2[x]/(x^3+x^2+x+1)$. They are (represented by) all polynomials of degree $2$ or less: $$ 0\\ 1\\ x\\ x+1\\ x^2\\ x^2+1\\ x^2+x\\ x^2+x+1 $$ Now, we are looking for functions from the generating set $\{1, x, y\}\subseteq \Bbb Z[x, y]$ to $\Bbb F_2[x]/(x^3+x^2+x+1)$ that will give rise to homomorphisms. It turns out that any function that sends $1\mapsto 1$ will give rise to a homomorphsim (this is a property of $\Bbb Z[x, y]$, and polynomial rings in general: the $x$ and $y$ do not interact more than just the bare minimum for a ring, so they may be sent wherever, independently, without any issues). Which means we are entirely free to choose where to send $x$ and where to send $y$.

There are $8$ choices for where such a function can send $x$, and $8$ choices for where such a function can send $y$. Which is to say, there are $8^2 = 64$ choices in total for where to send $x$ and $y$. Each such choice, coupled with $1\mapsto 1$ will give rise to a homomorphism, no two such choices will give rise to the same homomorphism, and any homomorphism stems from such a choice. Thus there are $64$ homomorhisms in total.

$\endgroup$
  • $\begingroup$ Arthur but answer is $2^6$ $\endgroup$ – TheStudent Nov 18 '19 at 9:57
  • $\begingroup$ lets see ,if i will find a mistake in solution . $\endgroup$ – TheStudent Nov 18 '19 at 9:58
  • 1
    $\begingroup$ @JyrkiLahtonen You're completely right. And it's not even morning here any more. I have no excuse. $\endgroup$ – Arthur Nov 18 '19 at 10:04
  • 1
    $\begingroup$ @TheStudent You were right, the answer turned out to be $2^6$. $\endgroup$ – Arthur Nov 18 '19 at 10:04
  • $\begingroup$ Thankyou @Arthur $\endgroup$ – TheStudent Nov 18 '19 at 10:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.