# Having difficulties showing this map is well defined?

Say $J$ is an ideal of $K[Y_1,...,Y_m]$ and $I$ of $K[X_1,...,X_n]$.
Then $\phi:K[Y_1,...,Y_m]/J \to K[X_1,...,X_n]/I$ defined by $Y_i \mapsto f_i, i\in\{1, \dots ,m\}$ is well defined $\iff$ $J=(g_1, \dots ,g_t)$, then, for each $i \in \{1, \dots t\}$, $g_i(f_1, \dots f_m) \in I$.

I have to use the Groebner basis of $I$ to do this but this trivial-looking result has so far been difficult to prove. Thanks you for your help.

• First use the universal property of polynomial rings to define a (well-defined) map $K[Y_1,\ldots, Y_m]\to K[X_1,\ldots, X_n]/I$, sending $Y_i\mapsto f_i$. Your condition precisely says $J$ is contained in the kernel of this map. Apr 30, 2016 at 14:02
• @Mohan, I don't understand, can you provide more details, please? Apr 30, 2016 at 14:15

Sorry for interchanging the roles of $K[X_1,\ldots,X_n]$ and $K[Y_1,\ldots,Y_m]$, but I am sure you will still get it.

$\textbf{Lemma:}$ Given a ring homomorphism $f\colon A\to B$ and elements $b_1,\ldots,b_n$ in $B$, there exists a unique homomorphism $F\colon A[X_1,\ldots,X_n]\to B$ sending $X_i$ to $b_i$ such that the restriction of $F$ to $A$ is just the map $f$.

This is not hard to prove, the statements tells you what $F$ must look like.

Applying this to $A=K$, $B=K[Y_1,\ldots,Y_m]$, $b_i=f_i$ and the inclusion $A \hookrightarrow B$, gives you a homomorphism $\varphi\colon K[X_1,\ldots,X_n]\to B$ whose composition with the projection $B\to B/I$ is the homomorphism $\Psi\colon K[X_1,\ldots,X_n]\to B/I$ sending $X_i$ to the residue class of $f_i$.

If $J$ is an ideal of $A':=K[X_1,\ldots,X_n]$, what does it mean for the map $A'/J \to B/I$, $a+J \mapsto \Psi(a)=\varphi(a)+I$, to be well-defined? It means, that the map does not depend on the choice of the representative $a$ of the residue class $a+J$ in $A'/J$, i.e. given any other representative $a'$ of $a+J$, you have to check that $\Psi(a)=\Psi(a')$. But this is equivalent to:
$$a-a'\in J \Rightarrow \varphi(a)-\varphi(a') \in I$$ But $\varphi(a)-\varphi(a')=\varphi(a-a')$. So, the condition is clearly satisfied if $\varphi(J) \subset I$. Conversely, choosing $a\in J$ arbitrary and $a'=0$, the condition tells you that $\varphi(J)\subset I$. This shows that proving well-definedness is equivalent to checking $J\subset \operatorname{Ker}(\Psi)$

Let's look at $\varphi$: Consider a polynomial $P(X_1,\ldots,X_n) \in A'$. By the definition of a ring homomorphism, it suffices to understand how $\varphi$ acts on elements in $K$ and on $X_i$. By construction, $\varphi$ is the identity on $K$ and sends $X_i$ to $f_i$, i.e. $P(X_1,\ldots,X_n) \mapsto P(f_1,\ldots,f_n)$.
Thus, well-definedness is equivalent to $P(f_1,\ldots,f_n) \in I$ whenever $P(X_1,\ldots,X_n) \in J$. Again, by the definition of ring homomorphisms and ideals, it suffices to check this on a set of generators of $J$ (which in this case can be assumed to be finite by Hilbert's basis theorem).