Showing $x\to cx+b$ is a ring homomorphism

Question

In the book Algebra by Hungerford, at page 167, at question 14, it is asked that

Let $R$ be a commutative ring with identity, and $c,b \in R$ with $c$ is unit.Show that the assignment $x \to cx + b$ is a unique automorphism of $R[x]$ that is the identity map on $R$.

However, consider $R = \mathbb{Z}$, $c = 1$, $b = 2$. Then since this assignment is a homomorphism, it must be true that $$x(x+1) \to x^2 + x + 2 = (x+1)(x+3) = x^2 + 4x + 3,$$ which is clearly not true.Therefore, considering the fact that this question is even asked in math overflow, and got answer, what is wrong in my argument ?

Answer 1

Everything is fine. You are transforming by $x \mapsto x+2$ and extending by linearity and distribution. Call the resulting map $\phi: \mathbb{Z}[x] \to \mathbb{Z}[x]$.

With your example you have $$ \phi(x(x+1)) = \phi(x^2+x) = (x+2)^2+(x+2)= x^2+5x+6. $$ On the other hand, $$ \phi(x)\phi(x+1)= (x+2)(x+3) = x^2+5x+6 $$ so yes, they are equal.

Answer 2

.You should try to prove that $\phi(x)\phi(x+1) = \phi(x(x+1))$.

Note that $\phi(x) = x+2$, $\phi(x+1) = x+3$, and their product is $x^2+5x+6$.

Next, $x(x+1) = x^2+x$, and $\phi(x^2+x) = (x+2)^2 + x+2 = x^2+4x+4+x+2 = x^2+5x+6$.

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What you did, instead, was find $x \times x+1$, and compare this with $\phi(x) \times \phi(x+1)$ (which was also computed wrongly).

Note that $x \to cx+b$ means that for each $f \in R[x]$ we have $f(x) \to f(cx+b)$. In other words, $f$ is sent to $g$, the polynomial formed by substituting every occurence of $x$ in $f$, by $cx+b$.

The other way of saying this, is that any homomorphism from $R[x] \to R[x]$ which is the identity on $R$, is specified uniquely by where $x$ goes, because of the above correspondence.