From Irreducible Polynomials to Simple Algebraic Extensions
Let \( m \) be a non-constant irreducible polynomial over a field \( K. \) We will show that there exists a simple algebraic extension of \( K. \)
Notation
This article follows standard mathematical notation, particularly from group theory and ring homomorphisms. For clarity, key notations are explicitly outlined below:
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The notation \( K[t] \) denotes the ring of polynomials in the indeterminate \( t \) over the field \( K. \)
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The notation \( \langle m \rangle \) denotes the ideal of \( K[t] \) generated by the polynomial \( m \in K[t], \) i.e., \[ \langle m \rangle = \{ fm : f \in K[t] \}. \]
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Let \( G \) be a group with the group operator \( + \) and a subgroup \( H. \) The left coset of \( H \) by \( a \) is defined as \[ a + H = \{ a + h : h \in H \}. \]
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For a ring homomorphism \( \phi : R_1 \to R_2, \) where \( 0_2 \) is the additive identity of \( R_2, \) the kernel of \( \phi \) is defined as \[ \operatorname{Ker}(\phi) = \{ x \in R_1 : \phi(x) = 0_2 \}. \]
Known Results
The following discussion relies heavily on concepts from group theory and field homomorphisms. Presenting every proof from first principles would be impractical, 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 discussion below depends on.
Proposition QF. If \( m \in K[t] \) is a non-constant irreducible polynomial (i.e., \( \partial m \gt 0 \)), then \( K[t]/\langle m \rangle \) is a field.
Proposition CSE. 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 CE. Let \( G \) be a group with the group operator \( +. \) Let \( H \) be a subgroup of \( G. \) Let \( a, b \in G. \) Then the \( a + H = b + H \) if and only if \( a - b \in H. \)
Proposition KOM. Let \( \phi \) be a ring homomorphism. Then \( \phi \) is a monomorphism if and only if \( \operatorname{Ker}(\phi) = {0} \) where \( 0 \) is the additive identity in \( K. \)
Proposition PD. If \( f, g \in K[t] \) and \( g \ne 0, \) 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 CO. For an additive subgroup \( S \) of a ring \( R, \) and \( a, b \in R, \) the addition and multiplication of cosets defined as \( (a + S) + (b + S) = (a + b) + S \) and \( (a + S) \cdot (b + S) = (a \cdot b) + S \) are well defined operations, i.e., they produce consistent results regardless of the choice of \( a \) and \( b. \) The \( \cdot \) symbol is often omitted to keep the notation compact.
Proof
The proof is rather long, so it has been split into multiple subsections. For the sake of convenience, we will use the notation \( I = \langle m \rangle. \)
Part 1: 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, 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 CSE. 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.
Part 2: Elements of \( K[t]/I \)
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 CE since \( (qm + r) - r = qm \in I. \) Therefore an arbitrary element of \( K[t]/I \) may be written as \[ (a_0 + \dots + a_{n-1} t^{n-1}) + I \] where \( a_0, \dots, a_{n-1} \in K. \) Using Proposition CO expression, we can expand the above expression as \[ (a_0 + I) + \dots + (a_{n-1} + I)(t + I)^{n-1}. \]
Part 3: Zero of \( m \)
Now there is no guarantee that there is 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 CO, 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 CSE and the fact that \( m \in \langle m \rangle = I. \) Since \( I \) is the additive identity element of \( K[t]/I, \) we can say that \( t + I \) is a zero of \( m'. \)