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

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.

• Define $\phi:\Bbb C[x]\to\Bbb C$ via $\phi(f)=f(a)$. – Lord Shark the Unknown Jul 22 '19 at 7:01
• Pointwise addition and multiplication of function? – gaurav saini Jul 22 '19 at 7:08

$$x-a$$ 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...

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 ?

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.