How to find the extension of an ideal

Question

I have a very elementary question and I would appreciate understanding.

Let $A$ be a commutative and unitary ring, and $I$ an ideal of this ring.

Considering the inclusion $i: A\longrightarrow A[X]$, I need to prove that $I^e=I[X]$, where $I^e$ is the extension of the ideal $I$.

I know this is true because I have sometimes used it in other problems, but I actually don't know the reasoning behind it.

Thanks! Any help or explanation would be appreciated.

Answer 1

To see the other inclusion, take any $f \in I[x]$, then $f = a_0 + a_1 x + \dots a_n x^n$ for some $a_j \in I$. Now notice that $a_j \in I$ and $x^j \in A[x]$, so we have $f \in \{a_1 f_1 + \dots + a_n f_n \mid a_j \in I, f_j \in A[x]\}$.

Answer 2

Alex Vong gave a good suggestion in the comments, but there's another way to see this. Hopefully this answer is helpful, but to be honest, your answer is missing context (by which I mean, you haven't told us what you've tried or what your thoughts are), so when I answer such questions, it's mostly for fun, since without context, I can't know what will be most useful to you.

Much of this answer is exposition. If you're familiar with all the material contained here, this argument is much shorter.

The extension of an ideal satisfies a universal property. Let $f:A\to B$ be a ring homomorphism, $I$ an ideal of $A$, then $I^e:=f(I)B$ is the smallest ideal of $B$ containing the image of $I$, which means that if $g:B\to C$ is any ring homomorphism such that $f(I)\subseteq \ker g$, we must have $I^e\subseteq \ker g$. $I^e$ is the unique ideal with this property, since we can take $g$ to be the quotient $B\to B/I^e$, and then if another ideal $J$ satisfied the property, we would have $f(I)\subseteq J\subseteq \ker g= I^e$, and $I^e$ is the smallest ideal of $B$ containing $I^e$, so $J=I^e$.

Now the first isomorphism theorem says that if $g:B\to C$ then $I^e\subseteq \ker g$ is equivalent to the existence of a map $\tilde{g} : B/I^e \to C$ such that $\tilde{g}\circ q = g$, where $q:B\to B/I^e$ is the quotient.

Combining these pieces of information, we get the following.

If $A$ is a commutative, unital ring, and $I\subseteq A$ is an ideal. Then the extension of $I$ along the structure map $\iota:A\to A[x]$ is the unique ideal $I^e$ such that for all maps $g$ from $A[x]\to C$ such that $I\subseteq \ker g$, there exists a map $\tilde{g}:A[x]/I^e \to C$ with $\tilde{g}\circ q = g$.

Now we recall the universal properties of polynomial rings. Maps $g:A[x]\to C$ correspond precisely to pairs $(g',c)$ with $g'$ a map $g' : A\to C$ and $c$ an element of $C$. Maps $g:A[x]\to C$ with $I\subseteq \ker g$ therefore correspond to pairs $(g',c)$ with $I\subseteq \ker g'$. But by the universal property of the quotient, these correspond precisely to pairs $(\bar{g},c)$ with $\bar{g}$ a map from $A/I\to C$ and $c\in C$. Thus maps from $A[x]\to C$ which annihilate $I$ correspond precisely to maps from $A/I[x]$ to $C$. Conveniently, we have a surjective map $A[x]\to A/I[x]$ whose kernel is precisely $I[x]$ (this part is obvious from the definition of the map from $A[x]\to A/I[x]$, which is reduce every coefficient mod $I$). Thus by the universal property (of sorts) of $I^e$, since $A[x]/I[x]$ has the property characterizing $I^e$, $I[x]=I^e$.