# Isomorphism between polynomial ring and ring

Let $$\psi:R[X_1,..,X_d] \rightarrow S$$ be a surjective ring homomorphism from polynomial ring $$R[X_1,..,X_d]$$ to ring $$S$$ with $$\operatorname{Ker}(\psi)=I=I_0R[X_1,..,X_d]$$, where $$I_0$$ is an ideal of $$R$$. Show that (1) $$I_0=\operatorname{Ker}(\psi|_R)$$ and (2) $$S\cong\psi(R)[X_1,..,X_d]$$.

For the first question: From $$I_0R[X_1,..,X_d] = \{i_0f:i_0 \in I_0,f \in R[X_1,..,X_d]\}$$ follows that $$\forall f \in R[X_1,..,X_d]: \psi(f) \neq 0$$ we have $$\psi(i_0f) = \psi(i_0)\psi(f) = 0$$, hence $$\psi(i_0) = 0, \forall i_0 \in I_0$$. It proves that $$I_0 \subseteq \operatorname{Ker}(\psi|_R)$$. Also $$\forall r \in \operatorname{Ker}(\psi|_R): \psi(r)=0$$, so $$\operatorname{Ker}(\psi|_R) \subseteq I_0$$, hence $$\operatorname{Ker}(\psi|_R) = I_0$$.

For the second question: isomorphism theorem states that $$R[X_1,..,X_d]/I \cong S$$, so maybe prove $$R[X_1,..,X_d]/I \cong \psi(R)[X_1,..,X_d]$$?

• What have you tried? MSE is not designed to do your HW for you! – bounceback Mar 14 at 17:57
• @bounceback check edits – DeuzharNickens Mar 14 at 18:31
• 1) $R \to \psi(R)$ is necessarily surjective, and if $k \in \text{ker}(\psi\vert_R)$ then $k \subset I$, but $k \subset R$ so the only possibility is that $k \subset I_0$. For the other direction, use that $I_0 \subset I$. 2) Convince yourself of why $$S \cong \psi(R)[X_1,\dots,X_d] \cong R/I_O[X_1,\dots,X_d] \cong R[X_0,\dots,X_d]/<I_0>$$ – bounceback Mar 14 at 18:50
• @AromaTheLoop: You do understand that $R$ is a subring of $R[x_1,\ldots,x_d]$, right? The unit of $R$ is also the unit of the polynomial ring. Just like the $1$ of $\mathbb{Z}$ is the one in $\mathbb{Z}[x_1,\ldots,x_d]$. I mean, you don't seem disturbed by "$I_0R[x_1,\ldots,x_d]$", so surely you understand that elements of $R$ are viewed as elements of $R[x_1,\ldots,x_d]$ as well. – Arturo Magidin Mar 14 at 20:20
• @AromaTheLoop: However you defined the polynomial ring, you must have proven that it contains an isomorphic copy of $R$ via a canonical embedding $i\colon R\hookrightarrow R[x_1,\ldots,x_d]$ that is a unital map of rings with identity (that is, it sends $1_R$ to $1_{R[x_1,\ldots,x_d]}$. We then identify $R$ with its image inside of the polynomial ring and simply consider $R$ as being contained as a subring of the polynomial ring. – Arturo Magidin Mar 14 at 22:18