From Irreducible Polynomials to Simple Algebraic Extensions

By Susam Pal on 07 May 2025 [draft]

Let \( m \) be a non-constant irreducible polynomial over a field \( K. \) We will construct a simple algebraic extension of \( K \) and show that it is isomorphic to the quotient ring \( K[t]/\langle m \rangle. \) We begin by reviewing key concepts from abstract field theory that are relevant to our discussion, including a few formal results. These ideas are then illustrated using the field of rational numbers as a concrete example.

Contents

Definitions

Familiarity with algebraic structures such as groups, rings, and fields is assumed in this article. Familiarity with ring homomorphisms is also assumed. This article follows standard definitions and mathematical notation from the field of algebraic structures and ring homomorphisms. For clarity, some definitions and notations are explicitly outlined below:

Known Results

In the following sections, we will discuss several proofs pertaining to field extensions. These proofs rely on a substantial body of knowledge, particularly from group theory and field homomorphisms. Presenting each argument from first principles would be impractical in a brief article like this, as it would require a comprehensive treatment of the subject. To maintain brevity, we take several well-established results as given. Below is a list of key results that the subsequent discussion relies upon.

Proposition CM. Let \( G \) be a group with the group operator \( + \) and a subgroup \( H. \) Then \( a + H = H \) if and only if \( a \in H. \)

Proposition CDM. Let \( G \) be a group with the group operator \( +. \) Let \( H \) be a subgroup of \( G. \) Let \( a, b \in G. \) Then \( a + H = b + H \) if and only if \( -a + b \in H. \)

Proposition FM. A field homomorphism is also a field monomorphism.

Proposition KOM. Let \( \phi : R_1 \to R_2 \) be a ring homomorphism. Then \( \phi \) is a monomorphism if and only if  \( \operatorname{Ker}(\phi) = {0_1} \) where \( 0_1 \) is the additive identity in \( R_1. \)

Proposition KII. Let \( \phi : R_1 \to R_ 2\) be a ring homomorphism. Then \( \operatorname{Ker}(\phi) \) is an ideal of \( R_1. \)

Proposition PRPID. Let \( K \) be a field. Then \( K[t] \) is a principal ideal domain, i.e., all ideals of \( K[t] \) are principal ideals. In other words, all ideals of \( K[t] \) have the form \( \langle f \rangle \) for some \( f \in K[t]. \)

Proposition ISF. Let \( \phi : K_1 \to K_2 \) be a field homomorphism. Then \( \operatorname{Im}(\phi) \) is a subfield of \( K_2. \)

Proposition PD. Let \( f, g \in K[t], \) where \( K \) is a field, such that \( g \ne 0. \) Then there exists a unique pair of polynomials \( q, r \in K[t] \) such that \( f = gq + r \) and \( \partial r \lt \partial g. \)

Proposition PHCF. Let \( f, g \in K[t], \) where \( K \) is a field, such that \( g \) is irreducible and \( g \ne 0 \) and \( g \nmid f, \) then the highest common factor of \( f \) and \( g \) is the unit polynomial \( 1 \) and there exist polynomials \( u, v \in K[t] \) such that \( uf + vg = 1. \)

Proposition QR. If \( I \) is an ideal of a unital commutative ring \( R, \) then the set of cosets of \( I \) forms a unital commutative ring under the operations \( (a + I) + (b + I) = (a + b) + I \) and \( (a + I) \cdot (b + I) = (a \cdot b) + I. \) These operations are well defined, i.e., they produce consistent results regardless of the choice of \( a \) and \( b. \) This ring of cosets is called the quotient ring and it is denoted by \( R/I. \)

Proposition FIT. For a ring homomorphism \( \phi : R_1 \to R_2, \) we have \( R_1 / \operatorname{Ker}(\phi) \cong \operatorname{Im}(\phi), \) i.e., the quotient ring \( R_1 / \operatorname{Ker}(\phi) \) is isomorphic to the image of \( \phi. \)

Proof

Even after taking the several results from the previous section for granted, the proof remains rather long, so it has been split into multiple subsections. For convenience, we will use the notation \( I = \langle m \rangle. \)

Part 1: Field \( K[t]/I \)

Let \( K \) be a field. We will show that if \( m \in K[t] \) is a non-constant irreducible polynomial, then \( K[t] / \langle m \rangle \) is a field.

By Proposition QR, we know that \( K[t]/I \) is a unital commutative ring. Therefore it possesses the following properties of commutative rings: associativity and commutativity of addition, the existence of an additive identity and additive inverses, associativity and commutativity of multiplication, existence of the multiplicative identity, and distributivity of multiplication over addition. Note that the additive identity and multiplicative identity of \( K[t]/I \) are \( I \) and \( 1 + I, \) respectively. Now, in order to show that \( K[t]/I \) is also a field, we need to prove the existence of multiplicative inverse for each \( x \in K[t]/I \) when \( x \ne I. \)

Let \( x \in K[t]/I \) such that it satisfies \( x \ne I. \) Since \( x \in K[t]/I, \) it must be of the form \( x = f + I \) where \( f \in K[t]. \) If \( f \in I, \) then \( f + I = I \) by Proposition CM. But \( f + I = x \ne I. \) Therefore \( f \notin I. \)

If \( m \mid f, \) then \( f \in \langle m \rangle = I. \) But we have shown that \( f \notin I. \) Thus \( m \nmid f. \) Then by Proposition PHCF, there exist polynomials \( u, v \in K[t] \) such that \[ um + vf = 1. \] Using Proposition QR, we get \[ (v + I)(f + I) = vf + I = (1 - um) + I = 1 + I. \] The last equality follows from Proposition CDM using the fact that \( -(1 - um) + 1 = um \in \langle m \rangle = I. \) Since the product of \( v + I \) and \( f + I \) is the multiplicative identity of \( K[t]/I, \) we conclude that \( v + I \) is the inverse of \( f + I = x. \) This proves that for every \( x \in K[t]/I \) satisfying \( x \ne I, \) there exists a multiplicative inverse \( x^{-1} \in K[t]/I. \) Therefore \( K[t]/I \) is a field.

Part 2: Monomorphism \( \phi : K \to K[t]/I \)

Let \( m \) be a non-constant irreducible polynomial in \( K[t]. \) By Proposition QF, we know that \( K[t]/I \) is a field. Now we define the function \begin{align*} \phi : K &\to K[t]/I, \\ a &\mapsto a + I, \end{align*} where \( I = \langle m \rangle. \) We will now show that \( \phi \) is a monomorphism. First, we show that this is a homomorphism. The proof for this is straightforward. Let \( a, b \in K. \) Then \begin{align*} \phi(a + b) &= (a + b) + I = (a + I) + (b + I) = \phi(a) + \phi(b), \\ \phi(a \cdot b) &= (a \cdot b) + I = (a + I) \cdot (b + I) = \phi(a) \cdot \phi(b), \\ \phi(1) &= 1 + I. \end{align*} Note that \( 1 + I \) is the multiplicative identity of \( K[t]/I. \) This proves that \( \phi \) is a homomorphism. Now using Proposition FM we conclude here that \( \phi \) is a monomorphism.

Alternate Proof of Monomorphism

Although we have shown above that \( \phi \) is a monomorphism, we will discuss a slightly longer, alternate proof here that too shows that \( \phi \) is a monomorphism. This proof is redundant since we already have a succinct proof above. But we will discuss it anyway because it is quite interesting. If you would rather not read this redundant proof, please skip ahead to the last paragraph of this subsection. The alternate proof now follows. We know that \( \phi \) is a homomorphism. This has already been shown in the prevoius subsection. Since \( I \) is the additive identity of \( K[t]/I, \) we get \begin{align*} \operatorname{Ker}(\phi) &= \{ a \in K : \phi(a) = I \} \\ &= \{ a \in K : a + I = I \} \\ &= \{ a \in K : a \in I \}. \end{align*} The final equality above follows from Proposition CM. There is a subtle distinction between \( a \in K \) and \( a \in I, \) though in most cases, this difference is inconsequential. The notation \( a \in K \) refers to an element of \( K, \) whereas \( a \in I \) represents the corresponding constant polynomial in \( K[t]. \) Strictly speaking, we could write the latter as \( at^0 \in I, \) but such explicitness is generally considered an overkill. However, for clarity, let us rewrite the above equation using this more precise notation as follows: \[ \operatorname{Ker}(\phi) = \{ a \in K : at^0 \in I \}. \] To elaborate, we are looking for elements \( a \) in \( K \) such that the corresponding constant polynomials \( at^0 \) belong to \( I. \) But \( I = \langle m \rangle = \{ fm : f \in K[t] \} \) where \( m \) is a non-constant polynomial. The only possible constant polynomial in \( I \) is the zero polynomial. Therefore \[ \operatorname{Ker}({\phi}) = \{ a \in K : at^0 = 0 t^0 \} = \{ 0 \} \] where the final \( 0 \) is the additive identity element of \( K, \) and so is the coefficient \( 0 \) in \( 0 t^0. \) Now by Proposition KOM, \( \phi \) is a monomorphism.

Field Extension

We know that \( K \) is a field. By part 1, \( K[t]/I \) is a field too. In the above two subsections, we have demonstrated that \( \phi : K \to K[t]/I \) is a monomorphism. Therefore by the definition of field extension, \( \phi : K \to K[t]/I \) is a field extension.

Part 3: Zero of \( m' \) in \( K[t]/I \)

There may not be a zero of \( m \) in \( K. \) For example consider \( m = t^3 - 2 \in \mathbb{Q}[t]. \) There is no element \( \alpha \in \mathbb{Q} \) such that \( m(\alpha) = 0. \) But remarkably, if we treat \( m \) as a polynomial in \( (K[t]/I)[t], \) then we can definitely find a zero of \( m \) in \( K[t]/I. \) For clarity, we will use the notation \( m' \) when we treat \( m \) as a polynomial in \( (K[t]/I)[t]. \) Let us first write the original polynomial \( m \in K[t] \) as \[ m = a_0 + \dots + a_n t^n \] where \( a_0, \dots, a_n \in K. \) The corresponding polynomial in \( (K[t]/I)[t] \) is \[ m' = (a_0 + I) + \dots + (a_n + I) t^n. \] Now note that \( t + I \in K[t]/I. \) Evaluating the polynomial \( m' \) at \( t + I \) and using Proposition QR, we get \begin{align*} m'(t + I) &= (a_0 + I) + \dots + (a_n + I)(t + I)^n \\ &= (a_0 + \dots + a_n t^n) + I \\ &= m + I \\ &= I. \end{align*} The last equality follows from Proposition CM and the fact that \( m \in \langle m \rangle = I. \) Since \( I \) is the additive identity element of \( K[t]/I, \) we conclude that \( t + I \) is a zero of \( m'. \)

Part 4: Ring Homomorphism \( K[t] \to K(x) \)

In this section, we will show that if \( K \to K(x) \) is a simple extension then the function \begin{align*} \theta : K[t] & \to K(x), \\ f & \mapsto f(x) \end{align*} is a ring homomorphism. Now for any \( f = a_0 + \dots + a_m t^m \in K[t] \) and \( g = b_0 + \dots + b_n t^n \in K[t], \) where \( m \le n, \) we get \begin{align*} \theta(f + g) &= \theta\left( (a_0 + b_0) + \dots + (a_m + b_m) t^m + \dots + (0 + b_n) t^n \right) \\ &= (a_0 + b_0) x + \dots + (a_m + b_m) x^m + \dots + (0 + b_n) x^n \\ &= (a_0 + \dots + a_m x^m) + (b_0 + \dots + b_n x^n) \\ &= f(x) + g(x) \\ &= \theta(f) + \theta(g), \\ \\ \theta(f \cdot g) &= \theta\left( (a_0 + \dots + a_m t^m) \cdot (b_0 + \dots b_n t^n) \right) \\ &= \theta\left( \sum_{i + j = 0} a_i b_j + \dots + \sum_{i + j = m} a_i b_j t^m + \dots + \sum_{i + j = m + n} a_i b_j t^{m+n} \right) \\ &= \sum_{i + j = 0} a_i b_j + \dots + \sum_{i + j = m} a_i b_j x^m + \dots + \sum_{i + j = m + n} a_i b_j x^{m+n} \\ &= (a_0 + \dots + a_m x^m) \cdot (b_0 + \dots b_n x^n) \\ &= f(x) \cdot g(x) \\ &= \theta(f) \cdot \theta(g), \\ \\ \theta(1) &= 1. \end{align*} Therefore \( \theta \) is a homomorphism. This result is not too interesting by itself. However, we will use this result in the next section to examine the kernel of \( \theta \) which will help us to complete the proof.

Part 5: Simple Algebraic Extension \( K \to K(\alpha) \)

Construct a symbol \( \alpha \) such that \( \alpha \) is a zero of \( m \in K[t]. \) Then we have a field extension \( K \to K(\alpha). \) By part 4, \begin{align*} \theta : K[t] &\to K(\alpha), \\ f &\mapsto f(\alpha) \end{align*} is a ring homomorphism. Therefore by Proposition KII, \( \operatorname{Ker}(\theta) \) is an ideal of \( K[t] \) and by Proposition PRPID, \( \operatorname{Ker}(\theta) = \langle f \rangle \) for some \( f \in K[t]. \) By definition, \( \operatorname{Ker}(\theta) = \{ h : \theta(h) = 0 \}. \) Therefore \[ \operatorname{Ker}(\theta) = \{ h : h(\alpha) = 0 \} = \{ fg : g \in K[t] \} \] for some \( f \in K[t]. \) Since \( m(\alpha) = 0, \) \( m \in \operatorname{Ker}(\theta). \) Therefore \( m = fc \) for some \( c \in K[t]. \) Since \( m \) is irreducible, \( \partial f = \partial m. \) Therefore \( c \) is a constant polynomial and \( f = mc^{-1}. \)

If \( x \in \langle f \rangle, \) then \( x = fg \) for some \( g \in K[t]. \) Then \( x = m c^{-1} g \in \langle m \rangle. \) Thus \( \langle f \rangle \subseteq \langle m \rangle. \) Similarly, if \( y \in \langle m \rangle, \) then \( y = mh \) for some \( h \in K[t]. \) Then \( y = fch \in \langle f \rangle. \) Thus \( \langle m \rangle \subseteq \langle f \rangle. \) Since \( \langle f \rangle \subseteq \langle m \rangle \) and \( \langle m \rangle \subseteq \langle f \rangle, \) we get \( \langle f \rangle = \langle m \rangle. \) Therefore \[ \operatorname{Ker}(\theta) = \langle m \rangle = I. \] Now by Proposition FIT, \[ K[t]/\operatorname{Ker}(\theta) \cong \operatorname{Im}(\theta). \] Thus \[ K[t]/I \cong \operatorname{Im}(\theta). \] This shows that \( \operatorname{Im}(\theta) \) is a field. By Proposition ISF, \( \operatorname{Im}(\theta) \) is a subfield of \( K(\alpha). \) Therefore \( \operatorname{Im}(\theta) \subseteq K(\alpha). \) Since \( \theta(k) = k \) for all \( k \in K \) and \( \theta(t) = \alpha, \) both \( K \) and \( \alpha \) are contained in \( \operatorname{Im}(\theta). \) Now \( K(\alpha) \subseteq K(\alpha) \cap \operatorname{Im}(\theta) = \operatorname{Im}(\theta). \) Since \( \operatorname{Im}(\theta) \subseteq K(\alpha) \) and \( K(\alpha) \subseteq \operatorname{Im}(\theta), \) we get \[ \operatorname{Im}(\theta) = K(\alpha). \] Therefore \[ K[t]/I \cong K(\alpha). \] From part 2, we know that \( K \to K[t]/I \) is a field extension. Since \( K \cong K \) and \( K[t]/I \cong K(\alpha), \) the field extensions \( K \to K[t]/I \) and \( K \to K(\alpha) \) are isomorphic.

Elements of \( K[t]/I \)

As before, we use the notation \( I = \langle m \rangle. \) Now we will show that every element of \( K[t]/I \) can be written in the form \( r + I \) where \( r \in K[t] \) and \( \partial r \lt \partial m. \) Consider an arbitrary element \( f + I \in K[t]/I \) where \( f \in K[t]. \) Since \( m \ne 0, \) by Proposition PD, we can write \[ f = qm + r \] in a unique way where \( q, r \in K[t] \) and \( \partial r \lt \partial m. \) Therefore \[ f + I = (qm + r) + I = r + I. \] The second equality above follows from Proposition CDM since \( -r + (qm + r) = qm \in I. \) Therefore an arbitrary element of \( K[t]/I \) may be written as \[ (a_0 + \dots + a_{n-1} t^{n-1}) + I \] where \( a_0, \dots, a_{n-1} \in K. \)

Illustration

What we have shown in the previous section is that given an irreducible polynomial \( m \in K[t] \) that does not have a zero in \( K, \) we can simply construct a new symbol and call that the zero of \( m. \) In the previous section, we used the symbol \( \alpha \) to represent that "imagined" zero. This symbol then lies in a larger field \( K(\alpha). \) Further the properties and behaviour of this \( \alpha \) are well-defined since \( K(\alpha) \) is isomorphic to \( K[t]/\langle f \rangle. \) In fact, we can define the isomorphism maps every \( g(\alpha) \in K(\alpha) \) to \( g(t) + \langle m \rangle \in K[t]/\langle m \rangle, \) and vice versa. As a specific case, the isomorphism maps \( \alpha \) to \( t + \langle m \rangle \) which we have already shown is the zero of \( m' \in K[t]/I. \) Therefore, the behaviour of the zero \( \alpha \) of \( m \) mimics the behaviour of the zero \( t + I \) of \( m'. \)

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