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Hi my question is as follows:

Let A be a domain, and let $A[T]$ be the ring polynomials over A.

Show that for any elements $a,b{\in}A$, there exists a unique ring homomorphism ${\varphi}_{a,b}:A[T]{\to}A[T]$ such that:

(i) ${\varphi}_{a,b}(T)=aT+b$ in $A[T]$ and

(ii) ${\varphi}_{a,b}$ induces the identity map on $A$.

I have thought of using the universal property of polynomials ring by first invoking a ring homomorphism ${\varphi}:A{\to}A[T]$ but this is where I am stuck. I have no idea how to idea how to construct such a map ${\varphi}$ that will allow me to have map ${\varphi}_{a,b}$ satisfying the mentioned hypothesis.

Any hints/help would be welcomed. Thanks!

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    $\begingroup$ Observe that (i) and (ii) define a homomorphism. Why is it unique? $\endgroup$ – Paul K Mar 17 '17 at 10:05
  • $\begingroup$ Am i right to say that because it is uniquely determined by coefficients a and b? $\endgroup$ – thedilated Mar 17 '17 at 10:10
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    $\begingroup$ Yes. Do you know why (i) and (ii) determine a homomorphism and why it is welldefined? $\endgroup$ – Paul K Mar 17 '17 at 10:12
  • $\begingroup$ From what I understand is that this map sends a polynomial p(T) to a scaled factor of T to p(aT+b). Then from here one can easily show that it satisfies the conditions of being a ring homomorphism and is welldefined. $\endgroup$ – thedilated Mar 17 '17 at 10:29
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Your ring homomorphism is an $A$-algebra homomorphism from $A[T]$ to itself. Now for any $A$-algebra $B$, an algebra homomorphism from $A[T]$ to $B$ is uniquely determined by the image of the indeterminate $T$ in $B$. In other words: $$\DeclareMathOperator{\Hom}{Hom}\Hom_{A\text{-alg}}(A[T],B)\simeq B,$$ and more generally, for any set $I$ $$\Hom_{A\text{-alg}}(A[T_i]_{i\in I},B)\simeq B^I$$ ($\simeq\;$ denotes here a bijection).

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