# Two Equivalent Notions of Algebraic Simple Extension (Proof)

When reading texts about field extensions I've come across the following two definitions for the simple extension $K(\alpha)$ where $\alpha$ some algebraic number over $K$. They are

(1) $K(\alpha)$ is the smallest field containing both $K$ and $\alpha$

(2) $K(\alpha)=\{\sum_{0}^{\infty}k_i\alpha^i:k_i\in K\}$

I haven't been able to prove that these two are equivalent. In particular I don't even see why the second is a field necessarily. I'm sure I understood this at some stage though! Could someone give me a hint?

Many thanks.

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The second set is the set of polynomial expressions in $\alpha$ and is usually denoted $K[\alpha]$. That it is the same as its field of fractions is a theorem, which is proved as in the answers already given. (Note also that you wrote the set of power series in $\alpha$. –  lhf May 14 '12 at 23:12

The second expression isn't really what's intended - in most situations in algebra, infinite sums don't make sense (one can consider Laurent series however). The correct expression is $$K(\alpha)=\{f(\alpha)\mid f\in K[x]\}$$ or equivalently $$K(\alpha)=\{\textstyle\sum_{i=0}^\infty k_i\alpha^i\mid k_i\in K, \text{ and almost all }k_i=0\}.$$ To show that the two definitions are equivalent, you just need to show that $$L=\{f(\alpha)\mid f\in K[x]\}$$ is the smallest field containing $K$ and $\alpha$. This amounts to showing that $L$ is a field, and that any field containing both $K$ and $\alpha$ will necessarily contain $L$.

It should be fairly clear that $L$ is a ring - we have $0$ and $1$, the sum of $f(\alpha)$ and $g(\alpha)$ is just $(f+g)(\alpha)$ where we certainly have $f+g\in K[x]$, and similarly with multiplication, etc. In fact, we can consider the evaluation homomorphism $$\text{ev}_\alpha:K[x]\to L$$ defined by $$\text{ev}_\alpha(f)=f(\alpha).$$ This is a surjective ring homomorphism, with kernel $$J=\ker(\text{ev}_\alpha)=\{f\in K[x]\mid f(\alpha)=0\}.$$ By the first isomorphism theorem, we have that $$K[x]/J\cong L.$$ By the assumption that $\alpha$ is algebraic, the kernel $J$ is non-trivial, because we know there is some non-zero polynomial with $\alpha$ as a root. In fact, $$J=(m_\alpha)$$ where $m_\alpha\in K[x]$ is the minimal polynomial for $\alpha$. We know that the minimal polynomial is irreducible as an element of $K[x]$, and thus $J$ is a prime ideal (because $K[x]$ is a UFD). In a PID, a non-zero prime ideal is maximal. Thus, $$L\cong K[x]/J$$ is a field.

Finally, note that any field containing $K$ and $\alpha$, being closed under addition and multiplication, has to contain $\alpha^n$ for all $n\in\mathbb{N}$, and therefore also contain $k\alpha^n$ for all $k\in K$ and $n\in\mathbb{N}$, and therefore also contain $$k_0+k_1\alpha+\cdots+k_n\alpha^n$$ for all $k_i\in K$ and $n\in\mathbb{N}$. These are the elements of $L$, so any field cotnaining $K$ and $\alpha$ must also contain $L$.

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Thank you very much - a lot clearer now. It was remembering that evaluation homomorphism stuff I was struggling with! –  Edward Hughes May 14 '12 at 23:13
No problem, glad to help! –  Zev Chonoles May 14 '12 at 23:17
To see that the 2nd set is a field, let $g(x)$ be a polynomial in $K[x]$ that is not divisible by the minimal polynomial of $\alpha$ over $K$ (which we'll denote by $p(x)$). Note that by division algorithm we may find $a(x), b(x)$ such that $a(x)g(x) + b(x)p(x) = 1$, and therefore $a(\alpha) g(\alpha) = 1$ (since $p(\alpha) = 0)$. Hence $g(\alpha)$ has a multiplicative inverse which is $a(\alpha)$ (and note that both $g(\alpha)$ and $a(\alpha)$ are elements of the second set). Hence the second set is a field (it is obviously a commutative ring with 1). Note that if a field contains both $K$ and $\alpha$, it must also contain all elements in the set 2. On the other hand, the set 2 obviously contains $K$ and $\alpha$.