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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?

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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.

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  • $\begingroup$ 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
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    $\begingroup$ @jdminer, it is surjective but not injective because $0$ and $X^2$ go to $0$. $\endgroup$ – lhf Feb 22 '18 at 12:26
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    $\begingroup$ Oh i see, that makes sense! Thanks :) $\endgroup$ – jdminer Feb 22 '18 at 12:29

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