Isomorphic Mapping Between $\mathbb{Z}_3 [x] / \langle x^2 + x + 2 \rangle$ and $\mathbb{Z}_3 [x] / \langle x^2 + 1\rangle$

Question

The fields $\mathbb{Z}_3 [x] / \langle x^2 + x + 2\rangle$ and $\mathbb{Z}_3 [x] / \langle x^2 + 1\rangle$ are both exactly of the form:

$\{0, 1, 2, x, 1 + x, 2 + x, 2x, 1 + 2x, 2 + 2x\}$.

And since both of these fields have a one-to-one correspondence, we can have an isomorphic mapping $f$ that sends each element to itself, right? So would stating the mapping:

$0 \rightarrow 0$, $1 \rightarrow 1$, .... etc. be sufficient to show an isomorphism? Do I also need to prove the homomorphic property separately? If so, how would I do that? I know there's a theorem that states two fields of the cardinality $p^n$ are isomorphic, but I need to show a specific mapping. Thanks.

Answer 1

The function sending each element to itself in the other field isn't an homomorphism, because the product isn't the same, for example, let $A=\mathbb{Z}_3 [x] / \langle x^2 + x + 2\rangle$ and $B=\mathbb{Z}_3 [x] / \langle x^2 + 1\rangle$. In $A$, $$x(x+1)=x^2+x=-2=1$$

but in $B$, you have that

$$x(x+1)=x^2+x=x+1$$

So the property that homomorphism preserves the product is not correct.

To build a correct homomorphism, see that $f\colon B\longrightarrow A$ sends $f(a+bx)=f(a)+f(b)f(x)$, and since $f(1_B)=1_A$, it must happen that $f(a)=f(1\cdot a)=a$ and $f(b)=b$, so you only need to see where goes $f(x)$

Since your $x_B\in B$ has the property that $x_B^2=-1=2$, you must send it to another element in $A$ that has that property, and that is

$(x_A^2)^2=(-x_A-2)^2=x_A^2+4x_A+4=(-x_a-2)+x_A+4=2$

So the homomorphism that sends $x_B\mapsto x_A^2$ and $1_A\mapsto 1_B$ will be your answer. Let's check it:

$$f((a+bx)+(c+dx))=f(a+c+bx+dx)=a+bf(x)+c+df(x)=f(a+bx)+f(c+dx)$$

In the other part,

$$f((a+bx)(c+dx))=f(ac+adx+bcx+bdx^2)=f(ac+adx+bcx+2bd)=$$ $$ac+adf(x)+bcf(x)+2bd=ac+adf(x)+bcf(x)+2bd$$ $$=ac+adx^2+bcx^2+x^4bd$$ $$=(a+bx^2)(c+dx^2)=f(a+bx)f(c+dx)$$

Answer 2

You want commutative diagram:

$\require{AMScd}$ \begin{CD} \Bbb Z_3[x]@>\varphi>> \Bbb Z_3[x]\\ @VVV @VVV\\ \Bbb Z_3[x]/\langle x^2+x+2\rangle@>\bar\varphi>> \Bbb Z_3[x]/\langle x^2+1\rangle \end{CD}

where $\varphi$ is a ring homomorphism such that $\varphi(x^2+x+2) \in \langle x^2+1\rangle$ so we could induce $\bar\varphi$ by first homomorphism theorem. All we have to decide is where to send $x$. In general, $\varphi(x)$ can be polynomial of arbitrary degree, but in our case, it is not a loss of generality to assume it is of degree $1$. (Can you see why?)

So, let $\varphi(x) = ax + b$. Then $\varphi(x^2+x+2) = a^2x^2+a(2b+1)x+(b^2+b+2)$ and it must be divisible by $x^2 + 1$, so we conclude that $a^2x^2+a(2b+1)x+(b^2+b+2) = a^2x^2 + a^2$, i.e.

\begin{align} a(2b+1) &= 0\\ b^2+b+2 &= a^2 \end{align}

which we solve in $\Bbb Z_3$ to get $b = 1$, $a=\pm 1$. This gives us two different $\bar\varphi$, one that maps $x$ to $x+1$ and the other that maps $x$ to $2x+1$, more concretely, $\bar\varphi_1(ax+b) = ax+(a+b)$ and $\bar\varphi_2(ax+b) = 2ax+(a+b)$. One could now use linear algebra to immediately verify that both are bijective (we have $\Bbb Z_3$-linear maps between two-dimensional $\Bbb Z_3$-vector spaces).