show the following homomorphism is an isomorphism

Question

pretty sure one-to-one correspondence means bijective. This question is from Undergraduate Commutative Algebra by Miles Reid. I have a feeling that this is wrong, but maybe my definition of 'extending' is wrong

Answer 1

The point is the following. Suppose you have a morphism $f:A\to B$. Consider the ring of polynomials $A[X]$. Then we can pick $b\in B$ and define $\hat f:A[X]\to B$ by declaring that $\hat f(a)=f(a)$ if $a\in A$; $X\to b$ and extending linearly, so that $$a_0+a_1x+\cdots+a_nx^n\mapsto f(a_0)+f(a_1) b+\cdots+f(a_n) b^n$$ Conversely, if $\hat f:A[X]\to B$ is a ring morphism with $b=\hat f(X)$, there is a unique ring morphism $f:A\to B$ such that $\hat f\iota =f$ where $\iota :A\to A[X]$ is the canonical inclusion, namely let $\hat f$ be $f\iota$! Hence, a ring morphism $\hat f:A[X]\to B$ is essentially a pair $(f:A\to B,b)\in\hom(A,B)\times |B|$, i.e. there is a bijection

$$\hom(A[X],B)\longleftrightarrow \hom(A,B)\times |B|\\ \hat f\mapsto (\hat f\iota,f(X))$$

ADD This can be used as done by Martin B. here to show with Yoneda's lemma that for $f\in A$ an element in a ring, $A_f\simeq A[y]/(1-fy)$ quite smoothly.

Answer 2

Yes, there is a bijection here, namely between $B$ and $$\left\{\psi:A[T]\to B\ :\ \psi|_A=\varphi\right\},$$ and one direction of this bijection is $\psi\ \mapsto\ \psi(T)$, it will admit an inverse as the element $T$ is 'freely adjoint' to the ring $A$.