Consider $\varphi :\mathbb Z_2[X] \to R$ where $R$ is ring:

$R:= \mathbb Z_2 \times \mathbb Z_2$

$(a,b)+_R(c,d):= (a+c,b+d)$

$(a,b) \times_R (c,d):= (ac, ad+bc)$

and the function is defined by $\varphi(\sum a_iX^i)= (a_0, a_1)$

I need to show that is a ring homomorphism, injective or surjective.

I tried showing that:

$f(a+_Sb)= f(a)+_R f(b)$ and $f(a\times_S b)= f(a)\times_R f(b)$

($+_S$ and $\times_S$ are the operators for $\mathbb Z_2[X]$, standard polynomial multiplication and addition)

The addition I could show to be valid but I think I might be doing something wrong for multiplication as its not adding up. Also how do I show if it is injective or surjective?


Let's try to make sense of the multiplication in $R$.

Suppose $R=\mathbb Z_2[\theta]$. Then $(a+b\theta)(c+d\theta)=ac+(ad+bc)\theta+bd\theta^ 2$. So $\theta^2=0$.

Therefore, $R = \mathbb Z_2[X]/(X^2)$ and $\varphi$ is the quotient map.

  • $\begingroup$ I noticed my mistake (forgot about the $\theta ^2$ becoming 0) in showing the multiplication but what about showing if its a bijection? $\endgroup$ – jdminer Feb 22 '18 at 12:22
  • 1
    $\begingroup$ @jdminer, it is surjective but not injective because $0$ and $X^2$ go to $0$. $\endgroup$ – lhf Feb 22 '18 at 12:26
  • 1
    $\begingroup$ Oh i see, that makes sense! Thanks :) $\endgroup$ – jdminer Feb 22 '18 at 12:29

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.