I've been studying the way to adjoin an element that satisfies some polynomial equation $p$ (meaning it's a root) to a commutative ring with unity $R$, which is done by considering the quotient space $R[x]/(p(x))$. To avoid any possible problems, let's consider the polynomials we'll talk about monic and of degree greater or equal to 1, so each polynomial from $R[x]$ has a representative of degree strictly less than $p$ and each constant element from $R$ is preserved.
Now I'm considering how to adjoin multiple elements, each one satisfying one polynomial equation. There are two ways of doing this: directly, and inductively. My real question is about the equivalence of the two (see the end), but I would like a fast-check of the two ways I got. For the sake of simplicity, let's consider the case of two elements satisfying polynomials $p$ and $q$ respectively:
- Directly: Consider $p(x)$ and $q(y)$ as elements of $R[x,y]$. Then to adjoin the two elements we can consider $R[x,y]/(p(x),q(y))$, since $p(\bar x)=\overline{p(x)},q(\bar y)=\overline{q(y)}\in(p(x),q(y))$.
- Inductively: First we add the elemet satisfying $p$ the same way as always; by considering $R[x]/(p(x))$. Now we add the other element to the ring $R[x]/(p(x))$ by doing the same; $\Big(\big(R[x]/(p(x))\big)[y]\Big)/(q(y))$.
I really think this is correct, but I wrote it anyway in case I'm missing something. Anything to add?
Now, my real question is whether $R[x,y]/(p(x),q(y))\simeq\Big(\big(R[x]/(p(x))\big)[y]\Big)/(q(y))$ or not. Of course, intuitively, it has to, but I'm struggling a little bit justifying it. Can I use somehow that $R[x,y]\simeq\big(R[x]\big)[y]$? How would you do it?
Thanks in advance.
Best Answer
We have a natural immersion $R[x]\hookrightarrow R[x,y]$. This gives the composite map $R[x]\hookrightarrow R[x,y]\to R[x,y]/(p(x))$, whose kernel is precisely $(p(x))$. This gives you an induced map $R[x]/(p(x))\to R[x,y](p(x))$.
From here, the universal property of the polynomial ring gives you a morphism $(R[x]/(p(x)))[y]\to R[x,y]/(p(x))$, given by sending $y$ to $y$. Now compose with the quotient map to $(R[x,y]/(p(x)))/(\overline{q(y)})$, which by the isomorphism theorems is isomorphic to $R[x,y]/(p(x),q(y))$. This gives you a morphism $$\left(\frac{R[x]}{(p(x))}\right)[y] \longrightarrow \frac{R[x,y]}{(p(x),q(y))}.$$ The kernel of this map is precisely $(q(y))$, so you get a one-to-one morphism from $\Bigl(\bigl(R[x]/(p(x))\bigr)[y]\Bigr)/(q(y))$ to $R[x,y]/(p(x),q(y))$.
Now it’s just a matter of making sure the map is surjective. This will follow because the map includes at least the (equivalence classes of) $R$, $x$, and $y$ in its image, which generate.