How to Determine Algebraic Elements as a Subring in Advanced Ring Theory

TinyProf's Guide to Algebraic Elements and Subrings

What We're Solving: We want to investigate whether the collection of all algebraic elements in a ring R (over a subring S) naturally forms a subring itself. This is a beautiful question that connects polynomial theory with ring structure!

The Approach: To prove something is a subring, we need to show it's closed under subtraction and multiplication, and contains the additive identity. The key insight is that algebraic elements have a special property - they satisfy polynomial equations with coefficients from our base ring. We'll explore whether this property is "inherited" by sums, differences, and products of algebraic elements.

Step-by-Step Solution:

Step 1: Set up the definitions Let S ⊆ R be commutative rings with the same unity 1. An element r ∈ R is algebraic over S if there exists a non-zero polynomial f(x) ∈ S[x] such that f(r) = 0. Let A = {all elements in R that are algebraic over S}.

Step 2: Check if 0 and 1 are in A

• 0 satisfies f(x) = x ∈ S[x], so 0 ∈ A ✓
• 1 satisfies f(x) = x - 1 ∈ S[x], so 1 ∈ A ✓

Step 3: Investigate closure under addition This is where it gets interesting! If α, β ∈ A, is α + β necessarily in A?

Let's say α satisfies p(x) = 0 and β satisfies q(x) = 0, where p, q ∈ S[x].

The element α + β lives in the ring extension S[α, β], which is finite-dimensional over S (since both α and β are algebraic). Any element in a finite-dimensional extension is algebraic!

Here's why: S[α, β] is spanned by finitely many elements {αⁱβʲ}. So α + β can be written as an S-linear combination of these basis elements, which means the powers (α + β)ⁿ eventually become linearly dependent, giving us a polynomial relation.

Step 4: Investigate closure under multiplication Similarly, if α satisfies p(x) = 0 and β satisfies q(x) = 0, then αβ ∈ S[α, β], which is finite-dimensional over S, so αβ is algebraic over S.

Step 5: Investigate closure under subtraction If α ∈ A, then -α is also algebraic (if p(x) annihilates α, then p(-x) annihilates -α). Combined with closure under addition, we get closure under subtraction.

Step 6: The crucial insight The key insight is that finite extensions preserve algebraicity. When we adjoin finitely many algebraic elements to S, we get a finite extension, and every element in a finite extension is algebraic.

The Answer: YES! The set A of all algebraic elements forms a subring of R. This works specifically because:

• We're working with commutative rings (commutativity is essential)
• They share the same unity
• The finite extension property ensures closure under ring operations

This beautiful result tells us that "being algebraic" is a robust property that's preserved under the basic ring operations!

Memory Tip: Think of algebraic elements as "well-behaved" elements that satisfy polynomial equations. Since polynomials play nicely with addition and multiplication, combinations of algebraic elements stay algebraic. The motto is: "Finite extensions of algebraic elements are still algebraic!" 🎯

Great question - this connects so many beautiful ideas in abstract algebra! Keep exploring these connections between polynomial relations and ring structure.