Is $K[X]/(f)$ an algebraic extension?

Question

If $K$ a field and $f\in K[X]$ is irreducible, is $K[X]/(f)$ always algebraic? Those fields look strange to me. Indeed, for example, $\mathbb C=\mathbb R[X]/(X^2+1)$, but $\mathbb C$ are number whereas element of $\mathbb R[X]/(X^2+1)$ seem to be polynomials. So I'm a little bit perturbed.

So for my exercise, I have to show that if $x\in K[X]/(f)$ then there is a polynomial $g(T)\in K[X]/(f)[T]$ s.t. $g(x)=0$ right? How?

Answer 1

Yes, for $K$ a field and $f\in K[X]$ for irreducible then $K[X]/(f)$ is a field and this field is an algebraic extension of $K$.

For $\mathbb R[X]/(X^2+1)$ the elements are in fact equivalence classes (so sets) of polynomials. (Recall congruence modulo $n$, for a similar situation.) However, from each class/element you can chose one representative which has degree at most $1$ (for any polynomial consider the remainder on polynomial division by $X^2 + 1$), so the elements of this field are the classes of $a + bX$ with $a,b \in \mathbb{R}$, thus they are parametrized by pairs of real numbers. The correspondence to $\mathbb{C}$ is given by $a+ b X \mapsto a+bi$. It still needs to be shown that this is well-defined and compatible with the operations. To appreciate this informally note that $X^2$ divided by $X^2 +1$ leaves rest $-1$ so $X^2$ is (in the same class as ) $-1$.

More generally, the elements of $K[X]/(f)$ can be thought of polynomial expressions of degree at most $\deg (f) -1$. The "variable" in these polynomial expression is precisely a root of $f$.

Yes, to show that the extension field $K[X]/(f)$ is algebraic over $K$ what you have to do would be to show that for each element $A \in K[X]/(f)$ there is a polynomial $f \in K [T]$ such that $f(A)$.

To do this effectively you can note that that the class of $X$ is a root of $f$, essentially by definition. Then, note that the sum and product of algebraic numbers is algebraic and everything can be written via sums and products of the class of $X$ and elements of $K$.

Finally, you can really think of $K[X]/(f)$ as $K[\alpha]$ where $\alpha$ is some root of $f$. The point of considering $K[X]/(f)$ is that in abstract it is not at all clear if there is a root of $f$ or from which other field it should be taken. Indeed, the construction $K[X]/(f)$ can be used to show that every polynomial over a field $K$ has a root in some extension field of $K$.

Answer 2

A very direct way to see that $K[X]/(f)$ is an algebraic extension of $K$ is as follows:

• First of all $K[X]/(f)$ is a field. Indeed, $K[X]$ is a ring and $(f)$ is a maximal ideal in it (because $f$ is irreducible), hence the quotient is a field
• $K[X]/(f)$ is an extension of $K$. Indeed, the canonical map $K\to K[X]/(f)$, $a\mapsto a+(f)$ is a ring homomorphism and as $K$ is a fiueld it can only have two possible kernels: $\{0\}$ or $K$. The latter case does not occur because certainly $f\nmid 1$
• The extension is algebraic. Indeed, field extensions are also vector spaces over the smaller field. Here, $K[X]/(f)$ is clearly spanned by $1+(f),X+(f),X^2+(f),\ldots,X^{\deg f-1}+(f)$, hence finite dimensional. Thus for any $\alpha\in K[X]/(f)$, the (infinitely many) powers of $\alpha$ are $K$-linearly dependent, and any such linear dependence relation shows that $\alpha$ is the root of a polynomial in $K[X]$.