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Let $R$ be a commutative ring. then $I\trianglelefteq R\implies R[x]\otimes_R(R/I)\cong R[x]/I[x]$ as rings.

Here $\trianglelefteq$ means ideal.

I tried to construct a map and prove it's a ring isomorphism, but it didn't work out. Here's what I tried:

Let $\alpha$ be defined s.t. $\alpha(f(x)\otimes (r+I))=(r+f(x))+I[x]$ where $f\in R[x]$, $r\in R$. then $\alpha:R[x]\otimes_R(R/I)\to R[x]/I[x]$.

Injectivity: $\alpha(f(x)\otimes (r+I))=I[x] \implies r+f(x)\in I[x]\implies r\in I \implies f(x)\otimes (r+I)=f(x)\otimes I = 0\implies \ker\alpha=\{0\}.$

Surjectivity:

$g(x)+I[x]\in R[x]/I[x] \implies g(x)=h(x)\cdot x+c$ for some $c\in R$, $h(x)\in R[x]$ $\implies \alpha((h(x)\cdot x)\otimes (c+I))=g(x)$.

Homomorphism: $\alpha([f_1(x)\otimes (r_1+I)]\cdot [f_2(x)\otimes (r_2+I)])=\alpha(f_1f_2\otimes(r_1r_2+I))=f_1f_2+r_1r_2+I$?

$\alpha([f_1(x)\otimes (r_1+I)]+[f_2(x)\otimes (r_2+I)])=?$

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  • $\begingroup$ These questions could be useful : math.stackexchange.com/questions/2098781/…, math.stackexchange.com/questions/175789/… I first thought the first one was a duplicate, but it only gives a module isomorphism, so I retracted my vote. $\endgroup$ – Arnaud D. Jun 27 '17 at 9:22
  • $\begingroup$ Actually the fact that the well known module isomorphism also respects the ring structure is built in the definition of the ring structure on $A\otimes_R B$, where $A,B$ are $R$-algebras. So this comes as close to a duplicate as possible... $\endgroup$ – MooS Jun 27 '17 at 9:25

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