Explaining field extensions

Question

Let $K$ be a field. I have seen this construction in lectures but am having difficulty understanding it. Let $P(x) \in K[x]$, an irreducible polynomial, whose "root" may not exist in $K$ which we call $\alpha$. Then, $$ \frac{K[X]}{\langle P(X) \rangle} $$ is a field, and a bigger one that we started with. Firstly, why is this a field? And secondly, the only construction that I know of starts with an element $\alpha$ in some bigger field of $K$, and then derives the minimal polynomial of $\alpha$. But in the construction above we are starting from an irreducible polynomial (which is supposed to be the minimal polynomial of $\alpha$ I suppose), and ending by somehow deriving a (sometimes) non-existent (in $K$) element $\alpha$. Is there a way to explain this clearly and rigorously? I'm just having a hard time getting my head around these constructions.

(A construction that goes "the other way around" might be: Prove that $p(x)$ is irreducible in $F[x]$)

Answer 1

There is a lemma which says that if $R$ is a commutative ring and $\mathfrak m$ a maximal ideal in $R$, then $R/\mathfrak m$ is a field. This follows from the correspondence between ideals of $R$ containing $\mathfrak m$ and ideals of $R/\mathfrak m$ (they are in bijection). And fields are exactly the commutative rings with two ideals (the two trivial ideals containing either just $0$, or the entire ring). This is because an ideal contains a unit if and only if it is already identical to the entire ring. So if the zero ideal and the entire ring are the only ideals, then every element except for $0$ must be a unit. But that's what a field is. And we can also note that if $\mathfrak m$ is maximal, then there are exactly two ideals in which it is contained: $\mathfrak m$ itself, and all of $R$. By the correspondence with ideals of $R/\mathfrak m$, we see that $R/\mathfrak m$ has only two ideals, and is thus a field.

So in your case, we have $R=K[X]$ and $\mathfrak m=\langle P\rangle$. If $\langle P\rangle$ is maximal, then $K[X]/\langle P\rangle$ is a field. Luckily, this is the case, since $K[X]$ is Euclidean, and thus a principal ideal domain, in which ideals generated by an irreducible element are maximal. So the ideal in question is indeed maximal, and thus the corresponding quotient a field.

Now a short calculation will show that we can embed $K$ into this field, let's call it $L$, and so we can view $P$ as a polynomial in $L[X]$, and with some calculations we can see that the equivalence class of $X$ is a root of the embedded polynomial. So we gain a field which contains the original one (in the sense that there is a natural embedding), as well as a root of the given polynomial (again in the sense that it's a root of the embedded polynomial).

Answer 2

Let us say that you want to "add" a root of some polynomial $P(T)$ to a (commutative) ring $A$ (where $P$ has coefficients in $A$). First, you need to add an element $X$. Then you want to add all the things you can form with it by taking sums and products with itself and elements of $A$. This is exactly the idea behind the construction of the ring $A[X]$ of polynomials. Now you do not want to add any element, but you want a root of $P$. That is, $X$ needs to satisfy the relation $P(X) = 0$. Well, you just have to impose this relation. And since you still want a ring, this means taking the quotient by the ideal $(P(X))$ generated by $P(X)$. The image $\bar X$ of $X$ in the quotient $A[X]/(P(X))$ (which is a ring containing $A$) is a root of $P$. Indeed, $P(\bar X)$ is the image of $P(X)$ by the ring morphism $A[X] \twoheadrightarrow A[X]/(P(X))$, which is zero by definition of the quotient.

Ok, now, let us talk about fields. The problem in my construction is that even if $A$ is a field, $A[X]$ is in general not a field. This comes from the fact that if $P$ is the product of two non-constant polynomials, say $P = QR$, then in $A[X]/(P(X))$, we have $Q(\bar X)R(\bar X) = P(\bar X) = 0$. But $Q$ and $R$ are not divisible by $P$, so they are not in the ideal $(P)$, which means that they neither $Q(\bar X)$ nor $R(\bar X)$ are equal to $0$ in $A[X]/(P(X))$ : they are zero divisors. So if you want $A[X]/(P(X))$ to be a field, you need $A$ to be a field, but you also need $P$ to be irreducible. And these conditions are sufficient for the quotient $A[X]/(P(X))$ to be a field: if $P$ is irreducible, then for any $Q$ not divisible by $P$ (that is, $Q(\bar X)$ is a non-zero element in the quotient), you can find a Bézout relation $UP + VQ = 1$, whose image in the quotient is $U(\bar X)Q(\bar X) = 1$ (that is, $Q(\bar X)$ is invertible in the quotient).

Answer 3

If $\bar{Q}$ is a non zero class of $K[X]/(P)$, then $P$ does not divide $Q$. Since $P$ is irreducible , $P$ and $Q$ are coprime so there exist polynomials $U,V$ such that $UQ+VP=1$. Then $\bar{U} \ \bar{Q}=\bar{1}$, hence $\bar{Q}$ is invertible, so $K[X]/(P)$ is a field.

Write $P=X^n+a_{n-1}X^{n-1}X^{n-1}+\cdots+a_1X+a_0$. Setting $\alpha=\bar{X}$ and using the laws on the quotient yields $\alpha^n+\bar{a}_{n-1}\alpha^{n-1}+\bar{a}_0=\overline{P(X)}=\bar{0}$, so $\alpha$ is a root of $T^n+\bar{a}_{n-1}T^{n-1}+\cdots+\bar{a}_0\in (K[X]/(P))[T]$

Since $\iota: K\to K[X]/(P)$ is injective , we may abusively identify $K$ with it image in the quotient, and you can consider that $K$ is contained in $K[X]/(P)$, and that $\alpha$ is a root of $P$.

If you don't want to do that, you need to perform a well-known set theoretic trick: set $L=K\cup (K[X]/(P)\setminus \iota(K)$, and note that the union is disjoint. Define a bijective map $f: L\to K[X]/(P)$ which is the identity on $K[X]/(P)\setminus\iota(K)$ and wich maps $\lambda\in K$ to $\iota(\lambda)=\bar{\lambda}.$ Then define laws on $L$ using $f$: for all $x,y\in L$, set $x+_L y=f^{-1}(f(x)+f(y))$ and $x\cdot _L y=f^{-1}(f(x)\cdot f(y))$. Then $L$ is a field (tedious but easy) containing $K$, and $\alpha\in L$ is a root of $P\in K[X]$ (for real, this time). It shouldn't be hat complicated to check that $L=K(\alpha)$.

Note that by definition of the laws on $L$ , $f$ is an isomorphism, so $K(\alpha)=L\simeq K[X]/(P)$.

For the other way round, if $\alpha$ is an element of an algebraic closure of $K$, and $P$ is the minimal polynomial of $\alpha$ over $K$, evaluation $K[X]\to L$ at $\alpha$ induces an isomorphism $K[X]/(P)\simeq K(\alpha)$ via the first isomorphism theorem.

Answer 4

Let $\beta$ be a nonzero element of $L=K[x]/(P(x))$. Then $\beta=f(x)+(P(x))$ for some $f(x)$ in $K[x]$, relatively prime to $P(x)$. Since $f(x)$ is relatively prime to $P(x)$, there are polynomials $g(x)$ and $h(x)$ in $K[x]$ such that $f(x)g(x)+P(x)h(x)=1$. In $L$, this equation is $(f(x)+(P(x)))(g(x)+(P(x)))=1+(P(x))$, so $\beta=f(x)+(P(x))$ is invertible, so $L$ is a field.

Now define $\alpha$ in $L$ by $\alpha=x+(P(x))$. Then $P(\alpha)=P(x+(P(x)))=P(x)+(P(x))=0$ in $L$, so $\alpha$ is a zero of $P(x)$ in $L$.

Answer 5

First of all, since $P(X)$ is irreducible, the ideal it generates is maximal in $K[X]$ (do you know something about ideal theory?), therefore $L:=K[X]/\langle P(X)\rangle$ is a field (see for example this: https://proofwiki.org/wiki/Maximal_Ideal_iff_Quotient_Ring_is_Field). Now call $\alpha$ the class of $X$ in the quotient $K[X]/\langle P(X)\rangle$. We claim that $\alpha$ is a root of $P(X)$ in the field $L$. This is actually an easy computation, from which we derive that $P(\alpha)=0$ in $L$. We have constructed a field $L$ containing a root of $P(X)$. We can say more: we have maps $$K\to K[X]\to K[X]/\langle P(X)\rangle=L\to K(\alpha),$$ where the composition $$K\to L$$ is injective (since it is a nonzero field homomorphism) and the last map $$L\to K(\alpha),$$ which we get from the natural valuation-in-$\alpha$ map $K[X]\to K(\alpha)$ is an isomorphism.