Let $\mathbb Z_{3}\left [ i \right ]=\left \{ a+bi\mid a,b \in \mathbb{Z}_{3} \right \}.$

Show that the field $\mathbb Z_{3}\left [ i \right ]$ is ring isomorphic to the field $\mathbb Z_{3}\left [ x \right ]/\left \langle x^{2}+1 \right \rangle$

Here's my thought process:

  • I need to first show that a ring Homomorphism exists from the ring $Z_{3}\left [ i \right ]$ to the Quotient ring.

  • Then Check that a bijection exists.

  • Would the First Isomorphism theorem for rings work here?

Am I correct?

  • 1
    $\begingroup$ If you know that $\mathbb Z_3[i]$ and $\mathbb Z_3[x]/\langle x^2 + 1 \rangle$ are fields, then it is enough to show that there is a (non-trivial) surjective homomorphism from one to the other. Any homomorphism between fields is either trivial or an embedding (injective) — since any ideal of a field is either the entire field or trivial, as you proved earlier. $\endgroup$ – M. Vinay May 30 '16 at 6:35
  • $\begingroup$ If you want to use the first isomorphism theorem, define a homomorphism from the ring $\mathbb Z_3[x]$ to the field $Z_3[i]$ and show that $\langle x^2 + 1 \rangle$ is the kernel. $\endgroup$ – M. Vinay May 30 '16 at 6:38

You are basically correct. If you find a homomorphism between the rings, you only need to show if that particular homomorphism is bijective. (This approach is quite standard and in no way restricted to your particular case.)

The first isomorphism theorem is now trivial, so it can be left out.

  • $\begingroup$ It helps if you write out what the elements of $\mathbb Z_3[x]/(x^2+1)$ look like. This will make things quite obvious. $\endgroup$ – gebruiker May 30 '16 at 7:23
  • $\begingroup$ I think I had it figured out. What I did was to take the inverse map of the elements in the codomain and claim surjection holds. $\endgroup$ – Mathematicing May 30 '16 at 7:24
  • $\begingroup$ To be honest, I need to see what you actually did to see if you did it correctly. What you need to look out for is this: If you "take the inverse map" then you are assuming that an inverse map (from the entire codomain to the domain) exists. This can only be the case when you have a bijection, which is what you want to prove. $\endgroup$ – gebruiker May 30 '16 at 8:37

Start with the map

$$\begin{align}\Bbb{Z}_3\left[x\right]&\to \Bbb{Z}_3\left[i\right]\\P&\to P(i)\end{align}$$

It is a ring morphism by definition of the substitution that is surjective. Indeed take any $a+bi\in \Bbb{Z}_3\left[i\right]$ it is the image of the polynomial $bx+a$.

The kernel of this morphism is the ideal generated by $x^2+1$ because any polynomial such that $p(i)=0$ has $i$ as a root and therefore $-i$ because it has integer (real) coefficients and is divisible by $x^2+1$ so we can factor and we get the isomorphism required.

  • 1
    $\begingroup$ This would be a fine answer to the question: "Find an insomorphism between $\mathbb Z_3[x]/(x^2+1) $ and $\mathbb Z_3[i]$." The OP however mentions a thought process an wishes to know if it is correct. That's an entirely different question and seems to call for a different answer, IMHO. $\endgroup$ – gebruiker May 30 '16 at 6:47

Consider the homomorphism $\varphi:\mathbb{Z}_3[x]\to \mathbb{Z}_3[i]$ with $$\varphi(x)=i \text{ and } \varphi(n)=n \text{ for all } n\in \mathbb{Z}_3.$$

Then $\varphi(x^2+1)=i^2+1=0.$ Also, if $\varphi(f(x))=0$ for some $f(x)\in \mathbb{Z}_3[X]$, then $f(\varphi(x))=0\Rightarrow f(i)=0\Rightarrow x^2+1|f(x)$.

Hence, $\ker{\varphi}=(x^2+1)$. Also, $\varphi$ surjective, since for any $a+bi$, we have $\varphi(a+bx)=a+bi$. Hence, $\mathbb{Z}_3[i]\cong \mathbb{Z}_3[x]/(x^2+1)$.

In general, let $K$ be a field, if $a$ is a root of some monic irreducible polynomial $f(x)\in K[x]$, then $K[a]\cong K[x]/(f(x))$. We can consider the homomorphism $\varphi:K[x]\to K[a]$ with $$\varphi(x)=a \text{ and } \varphi(n)=n \text{ for all } n\in K.$$

Then $\ker{\varphi}=(f(x))$. Here, $f(x)$ is called the minimal polynomial of $a$ over $K$.


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