$\frac{\mathbb{C[x]}}{\langle x-a\rangle}$ is isomorphic to which field?

Question

We know that every polynomial of degree one is irreducible over $\mathbb{C}$ that is $\langle x-a \rangle$ is maximal ideal in $\mathbb{C[x]}$, hence $$\frac{\mathbb{C[x]}}{\langle x-a\rangle}$$ is a field. But I can't get which field? Which homomorphic function exist between $\mathbb{C}[x]$ and the field. As from the $\frac{\mathbb{C[x]}}{\langle x-a\rangle}$ one thing is clear that kernal of that homomorphism is $\langle x-a \rangle$. I am confused here.

Please help.

Answer 1

$x-a$ has degree $1$. The only degree $1$ extension of $\Bbb C$ is $\Bbb C$.

Or, necessarily $a\in\Bbb C$. So we get $\dfrac {\Bbb C[x]}{(x-a)}\cong\Bbb C (a)\cong\Bbb C$.

Or, use the evaluation homomorphism $e_a:\Bbb C[x]\to\Bbb C$ given by $e_a(f)=f(a)$. Then $\operatorname {ker}e_a=(x-a)$. And clearly $e_a$ is surjective. So by the first isomorphism theorem for rings...

Answer 2

The field $\frac{\mathbb{C[x]}}{\langle x-a\rangle}$ is obtained by taking the ring of polynomials $\mathbb C[x]$ and by adding the rule stating that the polynomial $x-a$ is zero, that is the rule $x=a$. When one takes a polynomial $f(x)$ and makes use of this new rule, one obtains $f(a)\in \mathbb C$. From this, you may infer that the field is no other than the base field $\mathbb C$.

To prove it more formally, I advise you to look at the map $\frac{\mathbb{C[x]}}{\langle x-a\rangle}\rightarrow \mathbb C$ which sends the class of the polynomial $f(x)$ to $f(a)$. Is this map well defined ? Is it an isomorphism ?

Answer 3

The use of $\mathbb{C}$ is a red herring. More generally, if $F$ is any field, then $F[x]/\langle x-a\rangle\cong F$ (where of course $a\in F$).

Indeed, you can consider the map $\varphi_a\colon F[x]\to F$, $\varphi_a(f)=f(a)$ (evaluation at $a$), which is a surjective ring homomorphism. Therefore $$ F\cong F[x]\big/\!\ker\varphi_a $$ by the homomorphism theorems. Conclude by observing/proving that $f\in\ker\varphi_a$ if and only if $x-a$ divides $f$.

By the way, the result holds true with $F$ replaced by any commutative ring $R$, because division with remainder in $R[x]$ is possible whenever the divisor is monic.