Let R be a commutative ring

Let $R$ be a commutative ring, let "a" be a fixed element in $R$, and let $g$ be the function from $R[x]$ to $R$ that sends each polynomial $f(x)$ to the element $f(a)$ in $R$. Show that $g$ is a surjective homomorphism of rings. Can you figure out the kernel of this homomorphism?

Im not sure what the functions are to get started on the proof of homomorphism

• Sorry, I'm not clear what you're asking. You don't know what the functions are? Do you know what $R[x]$ and $R$ are? This map takes a polynomial to its evaluation at $a$. How would you check it's a homomorphism? – Eric Auld Oct 4 '15 at 19:37
• I am just not sure how to get started on this proof. The set R[x] can be defined as a0+a1x+…+anxn – user276828998 Oct 4 '15 at 19:43

The homomorphism (usually called "evaluation at $a$") sends:

$f(x) = c_0 + c_1x + \cdots + c_nx^n \mapsto c_0 + c_1a + \cdots + c_na^n = f(a)$.

For example, if $R = \Bbb Z$, and we have:

$f(x) = x^2 - 2x - 3$, then the "evaluation at $4$" homomorphism sends $f \mapsto 5$.

Note that $f(x) = (x + 1)(x - 3)$, and indeed, $5 = (4 + 1)(4 - 3)$. Perhaps this gives you a clue as to why the evaluation homomorphism preserves multiplication (factoring).

If we call this homomorphism $\phi_a$, then to show $\phi_a$ is a ring homomorphism, you need to prove that:

1. $f(a) + g(a) = (f+g)(a)$
2. $f(a)g(a) = (fg)(a)$.

You might also want to consider what the image of "constant polynomials" might be under $\phi_a$.

The function $f: R[x] \rightarrow R$ which sends $f(x)$ to $f(a)$ is often called the "evaluation at $a$", and it simply takes a polynomial in $R[x]$ and replaces the unknown variable with $a$, thus giving you a number in return. For instance, let $g(x) = 2x + 1$ and $a = 3$. The function $f$ applied to $g[x]$ would then result in $g(3) = 2 \cdot 3 + 1 = 7$ The clue for a problem like this is to not focus too heavily on what the function "is" (as you state you're unsure of what the functions are) and instead look at the properties of the function.

In order to show that $f$ is a surjective (also often called onto) homomorphism, you must first check if it is a homomorphism and then check if it's surjective. For a function to be a homomorphism, it must satisfy these two properties: $$(g + h)(a) = g(a) + h(a) \\ (g \cdot h)(a) = g(a)\cdot h(a)$$ for $g,h \in R[x]$ and $a \in R$. Let \begin{align}h(x) & = h_{0} + h_{1}x + h_{2}x^{2} + ... + h_{n}x^{n}\\ g(x) & = g_{0} + g_{1}x + g_{2}x^{2} + ... + g_{m}x^{m}\end{align} And now you can check if the two properties above holds for these two polynomials. Remember, that in the two properties, the addition and multiplication on the left hand side are computed for the polynomials in $R[x]$ before yo uevaluate at $a$. The addition and multiplication on the right hand side are computed for the elements in $R$, after you evaluate at $a$.

To check for surjectivity, consider $b \in R$ and see if you can find a polynomial $p(x) \in R[x]$ such that $p(a) = b$, because then any element $b \in R$ can be evaluated to by choosing the right polynomial.

To find the kernel, you need to consider which polynomials in $R[x]$ that are evaluated to $0$ when you evaluate at $a$. For instance, let $p(x) = x^{2} - 4$, $a = 2$ and $f$ be the function which evaluates at $a$. Now, apply $f$ to $p(x)$ to get $p(a) = p(2) = 2^{2} - 4 = 0$, thus $p(x)$ is evaluated to $0$ for this particular $a$, and thus lies in the Kernel

Hope this helps