It's not direct sum/product. It's just addition and multiplication of the elements in each embedded ring (and each embedded ring has basically the same type of element).
Basically, we're saying R is all the elements of embedded rings under the exact same operations as the embedded rings and every embedded ring has the same operations.
Okay. Then no, the set-theoretic union of subrings is not itself necessarily a ring. Take $\Bbb Q[\sqrt{2}]\cup \Bbb Q[\sqrt{3}]$ for example. Multiplying or adding $\sqrt{2}$ and $\sqrt{3}$ together doesn't work. However we can speak of the ring generated by this set...
(Instead of viewing the elements of our collection of rings as embedded rings, it's easier just to think of them as subrings. In this case subrings of eg $\Bbb Q[\sqrt{2},\sqrt{3}]$)
Can you clarify what $\mathbb{Q}[\sqrt{2}]$ and $\mathbb{Q}[\sqrt{3}]$ are? You mean the ring of rational numbers over $\sqrt{2}$ and $\sqrt{3}$, respectively?
$R[x]$ is the polynomial ring, in this case for example $\Bbb Q[\sqrt{2}]$ consists of all sums $a+b\sqrt{2}$ with $a,b$ rational. (because $\sqrt{2}\cdot\sqrt{2}\in\Bbb Q$, it only requires two rationals to designate an emelent of this ring)
Oh, I see what you mean by your example. Because you can get the element $\sqrt{6}$, you haven't formed a ring (since that's not in the union). Correct?
I was trying to show that $R[X]$, where $R[X]$ is the collection of polynomials over an infinite amount of symbols (but with every polynomial using only a finite amount of symbols), is a ring if $R$ is a ring because, for every polynomial $p(x)$, there corresponds a ring $R[x]$ where $x$ is the symbols contained in $p.$
(Sorry if I did not communicate that well. I am still learning.)
@Limitless If you define it with the direct limit (a categorical construction), yes. Personally I think it's so clear as to not even require justification. :P
@BenjaLim If $R_1\hookrightarrow R_2\hookrightarrow R_3\cdots$, then we take the disjoint union of them and then quotient it by identifying two elements if one can be obtained from the other by throwing it through the embeddings
@BenjaLim the union? how do you add/multiply two elements of two different rings? to do that you'd need to interpret the embeddings as inclusions of subrings. (which is of course permissible I suppose)
The "tensor product" of commutative rings $(R_i : i \in I)$ is the commutative ring $R$ such that $\textbf{CRing}(R, S) \cong \prod_{i \in I} \textbf{CRing}(R_i, S)$ naturally for all commutative rings $S$.
@BenjaLim, I thought that by showing how every polynomial of $R[x_1,\dots]$ can be associated with a ring $R[x]$ where $x$ is the set of symbols used in the polynomial, $R[x_1,\dots]$ is therefore a ring. Sorry if my reasoning is flawed, I'm just now learning this stuff.
@ZhenLin, so my argument is correct? I am unfamiliar with the direct limit (haven't even touched category theory yet), so I didn't fully understand what anon was arguing.
means you can view R[x] inside R[x,y] inside R[x,y,z] inside etc. etc., and all of these can be "glued together"; adding or multiplying the elements from any two of them will result in an element of some R[x,y,...] down the line.
Sorry if this is a silly question, but what exactly is the difference between commutative algebra and a regular algebra? Do we restrict the algebra to only objects which satisfy commutativity?
