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 $IS\cap R=I$? 
 A: 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
A: 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
