caelestia's math center

Galois exercises~

We start with the easy.

Q. Take a nonzero $f\in F[x]$, and let $K:=F(a)\vert F$ be an extension with $f(a)=0$. Then

\[[K:F]\leq\deg f,\quad\text{equality iff }f\text{: irreducible.}\]

Proof. Starting from scratch, the initial knowledge we have about finite extensions should include the following.

If $F$ is a field, then $F[X]$ is a PID. Thus,
If $f$ is irreducible, then it generates a maximal ideal of $F[X]$. So $F[X]/(f)$ is a field and obviously has dimension $\deg f$ over $F$.
Conversely, suppose we have an algebra $L$ over $F$ and an element $x\in L$. We shouldn’t be surprised that $x$ is integral/algebraic over $F$ if and only if the subalgebra $F[x]\subset L$ is finite dimensional over $F$.
We can find a polynomial equation for $x$ as follows. Consider the linear endomorphism $m_x$ of $F[x]$ induced by left multiplication by $x$. Then $(x-m_x)$ is an $F[x]$-module endomorphism of $F[x]$. Of course it is $0$. Therefore, $x$ is a root of the monic polynomial $\operatorname{char}_F(m_x,X)=\det(X-m_x)\in F[X]$ (of the same degree as $F[x]$). Clearly this argument works for any associative algebra over a commutative ring.
Moreover, every algebraic element over a field has a minimal polynomial $f_x$, since $F[x]$ is also an ED. Minimal $\Longleftrightarrow$ Irreducible.
Then $F[x]$ has the basis $(1,x,x^2,\cdots,x^{\deg f_x-1})$.

These are enough! The minimal polynomial $f_a$ divides $f$, and $[K:F]=\deg f_a$.

One thing these problems have taught me is the importance of Gauss Lemma. It is easy to prove though:

Lemma.(Gauss) If $R$ is a UFD and $f$, $g\in R[X]$ are nonzero primitive polynomials, then $fg$ is also primitive. Primitive means $\gcd$ of the coefficients is $1$.

Proof. We can reduce modulo $p$ for every prime $p\in R$. But $(p)$ is a prime ideal and $R/(p)$ is a integral domain. We want to prove that $(R/(p))[X]$ is also an integral domain, which is obvious.

Corollary. If $R$ is a UFD, then $R[X_1,\cdots,X_n]$ is also a UFD whose primes are precisely:

primes $p$ of $R$, and
primitive polynomials $f\in R[X_1,\cdots,X_n]$ that are irreducible in $F[X_1,\cdots,X_n]$, where $F:=\operatorname{Frac}(R)$.

Proof. WLOG let $n=1$. Since $F$ is a field, $F[X]$ is a PID. Any $g\in R[X]$ has a unique factorization:
\[g=cg_1^{\alpha_1}\cdots g_k^{\alpha_k}.\]
WLOG suppose the $g_i$’s have coefficients in $R$ and are primitive by throwing the constants into $c$. By Gauss Lemma, $g/c\in R[X]$ is also primitive. Therefore $c$ is just the $\gcd$ of coefficients of $g$.

The fact that these two classes are irreducibles is clear. The argument above shows that every $g$ has a unique factorization into these two classes, which concludes the proof.

Post date: 2024/11/23
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