In my problem, we have the following, $A:=k[x,t]/(x^2-t)$ as a $k[t]$ algebra, we have the following term $$A\otimes_{k[t]} k[t]/(t-a)$$ I knew the rule that $M\otimes_R R/I = M/IM$. But then I was stuck, looking at the solution we also have the following $$A/(t-a)A = \frac{k[x,t]/(t-x^2)}{(t-a)k[x,t]/(t-x^2)} = k[x]/(x^2-a)$$ which made this tensor product much easier to understand.
People say these make a lot more intuitive sense with algebraic geometry, but without referring to algebraic geometry, do you know where can I find some reference for this kind of result?
I knew there are the isomorphism theorems but I guess this is not enough. Thank you very much!
Edit, I will add another example from my notes, we want to consider $\mathbb{Z}[\sqrt{-5}]/(2, 1+\sqrt{-5})$, we can do the following calculation $$\mathbb{Z}[\sqrt{-5}]/(2, 1+\sqrt{-5}) = \mathbb{Z}[x]/(x^2+5, 1+x, 2) =\mathbb{F}_2[x]/(x^2+5, 1+x) \\ =\mathbb{F}_2[x]/(x^2+1, 1+x) = \mathbb{F}_2[x]/((x+1)^2, 1+x) = \mathbb{F}_2[x]/(1+x) = \mathbb{F}_2.$$
Proposition: Suppose that $R$ is a ring, and $I$ and $J$ are two ideals of $R$. Then quotient of $R$ by the ideal generated by both $I$ and $J$ is isomorphic to the quotient of $R/I$ by $J(R/I)$ (the ideal generated by the image of $J$ under the canonical map $R\to R/I$). That is, if $S = R/I$, we have $$ R/(I + J)\cong S/JS. $$
Proof: We have canonical homomorphisms $f : R\to R/I = S$ and $g : S\to S/JS$ given to us by the construction of quotient rings. I claim that the composition $\phi = g\circ f : R\to S/JS$ is surjective with kernel $I + J$. Then the proposition will follow from the first isomorphism theorem. Because $f$ and $g$ are quotient homomorphisms, they are surjective.
We only need now show that $\ker\phi = I + J$. Unwrapping all the definitions and maps, we have:
\begin{align*} \ker\phi &= \{r\in R\mid \phi(r) = 0_{S/JS}\}\\ &= \{r\in R\mid g\circ f(r) = (0 + I) + JS\}\\ &= \{r\in R\mid f(r) - (0 + I)\in JS\}\\ &= \{r\in R\mid f(r)\in JS\}\\ &= \{r\in R\mid r + I\in JS\}\\ &= \{r\in R\mid r + I = (j + I)(s + I)\textrm{ for some }j\in J, s\in R\}\\ &= \{r\in R\mid r + I = (js + I)\textrm{ for some }j\in J, s\in R\}\\ &= \{r\in R\mid r - js\in I\textrm{ for some }j\in J, s\in R\}\\ &= \{r\in R\mid r - j'\in I\textrm{ for some }j'\in J\}\\ &= \{r\in R\mid r - j = i\textrm{ for some }j\in J, i\in I\}\\ &= \{r\in R\mid r = i + j\textrm{ for some }j\in J, i\in I\}\\ &= I + J. \end{align*} Thus, $R/(I + J)\cong S/JS$ by the first isomorphism theorem for rings. $\square$
The results in your question follow because if $I = (r_1,\dots, r_n)$ and $J = (s_1, \dots, s_m)$, then $$I + J = (r_1,\dots, r_n, s_1,\dots s_m) = (r_1) + \dots + (r_n) + (s_1) + \dots + (s_m).$$ By induction, you may compute $R/(I + J)$ in this case by successively quotienting by the $r_i$ and $s_j$ in any order you choose. You may compute $R/I$, where $R = \Bbb Z[\sqrt{-5}]$ and $I = (2,1 + \sqrt{-5})$, by first writing $R = \Bbb Z[x]/(x^2 + 5)$, applying the isomorphism in reverse, and then quotienting by the generators of the relevant ideal in a new order.