Extension and contraction of ideals in polynomial rings

Suppose $I$ is an ideal in a polynomial ring $R=k[x,y]$. Let $\overline{k}$ be the algebraic closure of $k$ and let $S=\overline{k} [x,y]$. Then is $IS\cap R=I$?

All rings will be commutative.

Lemma 1 Let $K$ be a field. Let $L$ be a non-zero $K$-algebra. Let $N$ a $K$-module. Then the canonical homomorphism $N \rightarrow N\otimes_K L$ sending $x$ to $x\otimes 1$ is injective.

Proof: Suppose $x \neq 0$. There exists a basis of $N$ over $K$ containing $x$. Since $1 \neq 0$ in $L$, there exists a basis of $L$ over $K$ containing $1$. Hence there exists a basis of $N\otimes_K L$ over $K$ containing $x\otimes 1$. Hence $x\otimes 1 \neq 0$ QED

Lemma 2 Let $K$ be a field. Let $A$ a $K$-algebra. Let $L$ be a non-zero $K$-algebra. Let $B = A\otimes_K L$. Then the canonical homomorphism $M \rightarrow M\otimes_A B$ is injective for any $A$-module $M$.

Proof: $M\otimes_A B = M\otimes_K L$. Hence the assertion follows from Lemma 1. QED

Proposition Let $K$ be a field. Let $A$ be a $K$-algebra. Let $L$ be a non-zero $K$-algebra. Let $B = A\otimes_K L$. By Lemma 1, we can identify $A$ as a subring of $B$ by the canonical homomorphism $A \rightarrow A\otimes_K L$. Let $I$ be an ideal of $A$. Then $I = IB \cap A$

Proof: By Lemma 2, the canonical homomophism $A/I \rightarrow A/I\otimes_A B$ is injective. Since $A/I\otimes_A B = B/IB$, we are done. QED

Corollary Let $K$ be a field. Let $A = K[X_1,\dots, X_n]$ be a polynomial ring. Let $L$ be an extension field of $K$. Let $B = L[X_1,\dots, X_n]$. Let $I$ be an ideal of $A$. Then $I = IB \cap A$

Proof: Since $B = A\otimes_K L$, the assertion follows immediately from the proposition. QED

Lemma Let $K$ be a field. Let $L/K$ be an extension field. Let $K[X_1,\dots, X_n]$ and $L[X_1,\dots, X_n]$ be polynomial rings. Let ($\omega_i$) be a linear basis of $L$ over $K$. Then every element $f \in L[X_1,\dots, X_n]$ can be uniquely written as $f = \Sigma_i \omega_i f_i(X)$, where $f_i(X) \in K[X_1,\dots, X_n]$.

Proof: Let ($M_{\alpha}$) be the family of all the monomials of $K[X_1,\dots, X_n]$. Let $f = \Sigma_{\alpha} c_{\alpha}M_{\alpha}$, where $c_{\alpha} \in L$. Let $c_{\alpha} = \Sigma_i a_{\alpha i} \omega_i$, where $a_{\alpha i} \in K$. Then $f = \Sigma_{\alpha} \Sigma_i a_{\alpha i} \omega_i M_{\alpha} = \Sigma_i \Sigma_{\alpha} a_{\alpha i} M_{\alpha} \omega_i = \Sigma_i \omega_i f_i(X)$, where $f_i(X) = \Sigma_{\alpha} a_{\alpha i} M_{\alpha}$.

Next we prove the uniqueness. Suppose $\Sigma_i \omega_i f_i(X) = 0$, where $f_i(X) \in K[X_1,\dots, X_n]$. Suppose $f_i(X) = \Sigma_{\alpha} a_{\alpha i} M_{\alpha}$, where $a_{\alpha i} \in K$. Then $\Sigma_i \omega_i f_i(X) = \Sigma_i \Sigma_{\alpha} \omega_i a_{\alpha i} M_{\alpha} = \Sigma_{\alpha} \Sigma_i \omega_i a_{\alpha i} M_{\alpha} = 0$. Hence $\Sigma_i \omega_i a_{\alpha i} = 0$ for each $\alpha$. Hence $a_{\alpha i} = 0$. Hence $f_i(X) = 0$. QED

Proposition Let $K$ be a field. Let $A = K[X_1,\dots, X_n]$ be a polynomial ring. Let $L/K$ be an extension field. Let $B = L[X_1,\dots, X_n]$. Let $I$ be an ideal of $A$. Then $I = IB \cap A$

Proof: Let $f \in IB$. Let ($\omega_i$) be a linear basis of $L$ over $K$. We can assume that one of $\omega_i$, say $\omega_{i_0}$ is 1. Let $f_1,\dots,f_n$ be generators of $I$. We can write $f = \Sigma_k g_k f_k$, where $g_k \in B$. Let ($M_{\alpha}$) be the family of all the monomials of $K[X_1,\dots, X_n]$. Suppose $g_k = \Sigma_{\alpha} c_{\alpha k} M_{\alpha}$, where $c_{\alpha k} \in L$. Then $f = \Sigma_k g_k f_k = \Sigma_k \Sigma_{\alpha} c_{\alpha k} M_{\alpha} f_k$. Since $M_{\alpha} f_k \in I$, we can write $f = \Sigma_i \omega_i h_i$, where $h_i \in I$. Suppose $f \in IB \cap A$. By Lemma, $f = h_{i_0}$. Hence $f \in I$. QED