# Algebriac Geometry - Morphisms of Algebraic Sets

1. Nov 1, 2013

### Math Amateur

I am reading Dummit and Foote (D&F) Section 15.1 on Affine Algebraic Sets.

On page 662 (see attached) D&F define a morphism or polynomial map of algebraic sets as follows:

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Definition. A map $\phi \ : V \rightarrow W$ is called a morphism (or polynomial map or regular map) of algebraic sets if

there are polynomials ${\phi}_1, {\phi}_2, .......... , {\phi}_m \in k[x_1, x_2, ... ... x_n]$ such that

$\phi(( a_1, a_2, ... a_n)) = ( {\phi}_1 ( a_1, a_2, ... a_n) , {\phi}_2 ( a_1, a_2, ... a_n), ... ... ... {\phi}_m ( a_1, a_2, ... a_n))$

for all $( a_1, a_2, ... a_n) \in V$

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D&F then go on to define a map between the quotient rings k[W] and k[V] as follows: (see attachment page 662)

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Suppose F is a polynomial in $k[x_1, x_2, ... ... x_n]$.

Then $F \circ \phi = F({\phi}_1, {\phi}_2, .......... , {\phi}_m)$ is a polynomial in $k[x_1, x_2, ... ... x_n]$

since ${\phi}_1, {\phi}_2, .......... , {\phi}_m$ are polynomials in $x_1, x_2, ... ... x_n$.

If $F \in \mathcal{I}(W)$, then $F \circ \phi (( a_1, a_2, ... a_n)) = 0$ for every $( a_1, a_2, ... a_n) \in V$

since $\phi (( a_1, a_2, ... a_n)) \in W$.

Thus $F \circ \phi \in \mathcal{I}(V)$

It follows that $\phi$ induces a well defined map from the quotient ring $k[x_1, x_2, ... ... x_n]/\mathcal{I}(W)$

to the quotient ring $k[x_1, x_2, ... ... x_n]/\mathcal{I}(V)$ :

$\widetilde{\phi} \ : \ k[W] \rightarrow k[V]$

$f \rightarrow f \circ \phi$

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My problem is, how exactly does it follow (and why?) that $\phi$ induces a well defined map from the quotient ring $k[x_1, x_2, ... ... x_n]/\mathcal{I}(W)$ to the quotient ring $k[x_1, x_2, ... ... x_n]/\mathcal{I}(V)$ ?

Can someone (explicitly) show me the logic of this - why exactly does it follow?

Peter

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2. Nov 2, 2013

### R136a1

In general: Let $R$ and $R^\prime$ be rings and let $I$ and $I^\prime$ be ideals of the respective rings. Let $f:R\rightarrow R^\prime$ be a ring morphism.

Let $\pi^\prime:R^\prime\rightarrow R^\prime/I^\prime$ be the canonical quotient morphism which sends an element $y$ in $R^\prime$ to its equivalence class $[y]$. So, we consider the map $\pi^\prime \circ f$ which sends $x\in R$ to $[f(x)]$.

Now, assume that for all $x\in I$, we have that $f(x)\in I^\prime$. This means that $\pi(f(x)) = 0$ for each $x\in I$. So $I\subseteq ker(\pi \circ f)$. This implies directly that there is a unique morphism $F:R/I\rightarrow R^\prime /I^\prime$ such that $F( [x]) = [f(x)]$. Let us prove this. Uniqueness is clear since I have defined $F$ at each point $R/I$ explicitely. To prove existence, assume that $[x] = [x^\prime]$ for some $x,x^\prime \in R^\prime$. Then $x-x^\prime \in I$, and thus $\pi(f(x-x^\prime)) = 0$. Thus $F([x]) = \pi(f(x)) = \pi(f(x^\prime)) = F([x^\prime])$.

3. Nov 3, 2013