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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] $?

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    $\begingroup$ What have you tried? MSE is not designed to do your HW for you! $\endgroup$ – bounceback Mar 14 at 17:57
  • $\begingroup$ @bounceback check edits $\endgroup$ – DeuzharNickens Mar 14 at 18:31
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    $\begingroup$ 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 $$S \cong \psi(R)[X_1,\dots,X_d] \cong R/I_O[X_1,\dots,X_d] \cong R[X_0,\dots,X_d]/<I_0>$$ $\endgroup$ – bounceback Mar 14 at 18:50
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    $\begingroup$ @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. $\endgroup$ – Arturo Magidin Mar 14 at 20:20
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    $\begingroup$ @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. $\endgroup$ – Arturo Magidin Mar 14 at 22:18

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