Isomorphism of rings between 2 non fields

140 Views Asked by At

Good day, I found a question which finds isomorphisms between 2 fields

Find an explicit isomorphism of rings

I am, however, interested in finding out how to construct such an isomorphism if the domain and codomain weren't fields. Notably,

$\lambda:\mathbb{F}_5[x]/(x^2+x+3)\rightarrow\mathbb{F}_5[x]/(x^2+3x+2)$

In lectures, we were told a little about this and were hinted to use idempotents but I am not quite clear on how exactly I am supposed to use them. Can I apply a similar technique as in the question that I linked? Except I cant seem to make sense of how the author of the answer got the relations for $a$ and $b$.

1

There are 1 best solutions below

0
On BEST ANSWER

What I found, is that using this rather universal method, such problems can be solved for fields and non-fields

Define a function $\lambda:\mathbb{F}_5[x]/(x^2+x+3)\rightarrow\mathbb{F}_5[x]/(x^2+3x+2)$ s.t.

$\lambda (x)=ax+b$

$\lambda(0)=0$

$\lambda(a)=a$

We can send the polynomial in the codomain to $\equiv0$ $mod(\text{polynomial in codmain})$ (notation is probably wrong but the idea will be clear below)

$\lambda(x^2+x+3)\equiv 0 \ mod(x^2+3x+2)$. Which means

$$(ax+b)^2+(ax+b)+3 = a^2(x^2+3x+2)$$

Should get something like $$2ab+a=3a^2 \text{ and } b^2+b+3=2a^2$$

Just plug in elements of $\Bbb F_5$ to see which ones solve the equations. I got, say, $a=3, b=4$

So, a possible isomorphism is $$\lambda(x)= 3x+4$$ All that is left is to show injectivity, surjectivity and the fact that this is a homomorphism.

It should all check out.