MHB Proof: $K[a]$ is a Field & if $K(a)=K[a]$, then $a$ is Algebraic Over $K$

[mathmari]

Hey!

$K \leq L, a \in L$

I am looking at the proof that if $a$ is algebraic over $K$, then $K(a)=K[a]$.

We show that $K[a]$ is a field, then we have that $K \subseteq K[a] \subseteq K(a) \subseteq L$.

Let $0 \neq c \in K[a]$, then $c=f(a), f \in K[x]$.
Let $p(x)=Irr(a,K)$.
Since $p(a)=0$ and $f(a) \neq 0$, we have that $p(x) \nmid f(x)$, so we have that $(p(x), f(x))=1$.
Therefore, there are $h(x), g(x) \in K[x]$ with $h(x) \cdot p(x)+g(x) \cdot f(x)=1$.
For $x=a$: $h(a) \cdot p(a)+g(a) \cdot f(a)=1 \Rightarrow h(a) \cdot 0+g(a) \cdot f(a)=1 \Rightarrow g(a) \cdot f(a)=1$.
We have the following:
$K \leq L, a \in L$

$K[a]=\{f(a), \text{ with } f(x) \in K[x]\}$
$K(a)=\{f(a) \cdot g^{-1}(a), \ \ f(x), g(x) \in K[x], g(a) \neq 0 \}$

Why does it stand that $$K \subseteq K[a] \subseteq K(a) \subseteq L$$ ??Could you explain me the proof above?? How did we show that $K(a)=K[a]$?? (Worried)

Does the reverse also stand?? Does it stand that if $K(a)=K[a]$, then $a$ is algebraic over $K$?? (Wondering)

 

[Deveno]

There are different definitions people start with.

$K(a)$ usually means the field generated by $K$ and $\{a\}$, in the following sense:

If $E$ is a field such that $K \subseteq E$ and $a \in E$, then $K(a) \subseteq E$.

$K[a]$ typically means the RING generated by $K$ and $\{a\}$, in an analogous fashion.

We have the standard "evaluation map": $\phi_a:K[x] \to K[a]$ given by:

$\phi_a(f(x)) = f(a)$. Since $x \in K[x]$, we have $a = \phi_a(x) \in \phi_a(K[x])$,

and certainly $K \subseteq K[a]$, so $\phi_a$ is onto.

It's not hard to show that no matter which $a \in L$ we choose, $\phi_a$ is a ring homomorphism.

(Basically we just write "polynomials in $a$" instead of "polynomials in $x$". Of course, in the larger field $L$ that $K[a]$ lives in, we may be able to simplify these).

If $a$ is algebraic, that means for SOME $p(x) \neq 0 \in K[x]$, we have $p(a) = 0$.

Another way to say this, is that $p(x) \in \text{ker }\phi_a$, so this homomorphism has a non-trivial kernel.

Now, we can proceed a couple of different ways, from this point. We can use the fact that $K[x]$ is a Euclidean domain (it has a division algorithm) to establish that the kernel of $\phi_a$ is generated by a single element, which must be irreducible. This, in turn, tells us the kernel is a maximal ideal in $K[x]$, and thus that the quotient ring $K[x]/(\text{ker }\phi_a) \cong K[a]$ is thus a field. Since $K[a]$ is a minimal ring, it must therefore be the minimal field $K(a)$, since it is a field.

This is a bit "abstract", and many people "don't get it".

Or, we can, as your text does, show that any non-zero $\phi_a(f(x)) = f(a)$ is invertible in $K[a]$. This also uses the fact that $K[x]$ is a Euclidean domain, where we assert we can find:

$g(x),h(x) \in K[x]$ such that $h(x)p(x) + g(x)f(x) = 1$

given that $\text{gcd}(f(x),p(x)) = 1$.

Basically, this shows that $g(a) = \dfrac{1}{f(a)}$, that is, every non-zero $f(a) \in K[a]$ is a unit.

The polynomial $p(x)$ is often taken to be MONIC, and is then called the minimal polynomial for $a$.

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Now, you may have seen a definition of $K(a)$ as:

$Q(K[a]) = \left\{\dfrac{f(a)}{g(a)}: f(a) \in K[a],g(a) \in K[a]^{\ast}\right\}$.

Strictly speaking, this is incorrect, because it's "too big", we have to take equivalence classes under the equivalence relation:

$\dfrac{f(a)}{g(a)} \sim \dfrac{h(a)}{k(a)} \iff f(a)k(a) = g(a)h(a)$

This is known as the "field of fractions" of $K[a]$.

We can include $K[a]$ in $Q(K[a])$ for every $f(a) \in K[a]$ as the equivalence class of $\dfrac{f(a)}{1}$. This mapping is then an injective ring-homomorphism:

$K[a] \to Q(K[a])$, so the image of $K[a]$ is an isomorphic ring to $K[a]$. This essentially allows us to "cancel common factors" in the numerator and denominator, the image of $K[a]$ is comprised of those "fractions" that after said cancellation, have a denominator of 1.

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About your second question: does the converse hold?

Suppose $a \in L$ is such that $K[a]$ is a field. Recall that $K[a] \cong K[x]/(\text{ker }\phi_a)$.

Note that $a$ is invertible in $L$, while $x$ is NOT invertible in $K[x]$. This tells us $\text{ker }\phi_a$ cannot be trivial, for if it was, we would have $K[a] \cong K[x]/(0) = K[x]$, an isomorphism, in which case $\phi_a$ would be invertible (and also a ring-homomorphism), and $x$ would thus be a unit of $K[x]$.

But if $\text{ker }phi_a \neq (0)$, there must be some non-zero polynomial IN it, say $k(x)$.

Thus $\phi_a(k(x)) = k(a) = 0$, that is: $a$ is algebraic.

In short: an extension ring $K[a]$ of $K$ made by adjoining $a \in L$ that is algebraic over $K$ is a field. For example:

$\Bbb Q[\sqrt{2}]$ is a field, since $\sqrt{2}$ satisfies the polynomial $x^2 - 2 \in \Bbb Q[x]$. In fact, given:

$a + b\sqrt{2} \in \Bbb Q[\sqrt{2}]$, with $a^2 + b^2 \neq 0$ we can compute explicitly that:

$\dfrac{1}{a + b\sqrt{2}} = \dfrac{a}{a^2 - 2b^2} - \dfrac{b}{a^2 - 2b^2}\sqrt{2} \in \Bbb Q[\sqrt{2}]$

(the only "tricky part" is showing the denominator $a^2 - 2b^2$ is never 0 for any rational numbers $a,b$).

 

[mathbalarka]

No point reinventing the wheel : have a look at my answer in http://mathhelpboards.com/linear-abstract-algebra-14/rationals-adjoin-cube-root-3-field-11965.html#post57042