$(a,b)$ is algebraically independent over $K$ iff $a$ is algebraically independent over $K$ and $b$ is algebraically independent over $K(a)$

Question

I’m using the following definition of algebraic independence:

If $K \subset L$ is a field extension, and $(\alpha_i)_{i\in I}$ is a subset of elements of $L$, we say that the subset $(\alpha_i)_{i\in I}$ is algebraically independent if the $K$-algebras homomorphism $K[(x_i)_{i\in I}]\rightarrow L$ such that $x_i \mapsto \alpha_i$ is injective.

Here, if I understand correctly, $K[(x_i)_{i\in I}]$ is the polynomial ring over $(x_i)_{i\in I}$. Please correct me if I’m wrong.

I also know that $(a,b)$ being algebraically independent over $K$ doesn’t follow in general from $a$ and $b$ being algebraically independent over $K$ (counterexemple taken from Wikipedia: $a=\sqrt{\pi}$, $b=2\pi+1$). However, I’d like to prove that $(a,b)$ is algebraically independent over $K$ iff $a$ is algebraically independent over $K$ and $b$ is algebraically independent over $K(a)$ (the statement is given without proof on my lecture notes).

I guess if $(a,b)$ is algebraically independent over $K$, then this means by definition that the morphism $K[x_1,x_2] \rightarrow L$ is injective, and then it’s restriction $K[x_1] \rightarrow L$ is injective too. Then $a$ is algebraically independent over $K$. We should see now that the morphism $K(a)[x_2] \rightarrow L$ such that $x_2 \mapsto b$ is injective too, but I’m not really sure how to do that.

For the other direction, I don’t really know what to attempt. Any help would be appreciated.

Answer 1

"the morphism $K(a)[x_2]\to L$ such that $x_2 \mapsto b$ is injective too" because its kernel is $$\{P(a,x_2)/Q(a)\mid Q\in K[x_1],Q\ne0, P\in K[x_1,x_2],P(a,b)=0\}$$$$=\{P(a,x_2)/Q(a)\mid Q\in K[x_1],Q\ne0,P=0\}=\{0\}.$$

For the other direction, assume that both morphisms $K[x]\to L,P(x)\mapsto P(a)$ and $K(a)[y]\to L,P(a,y)/Q(a)\mapsto P(a,b)/Q(a)$ are injective, i.e.:

1. the only polynomial $P(x)\in K[x]$ s.t. $P(a)=0$ is $P(x)=0$ and
2. every polynomial $P(x,y)\in K[x,y]$ s.t. $P(a,b)=0$ satisfies: $P(a,y)$ is the $0$ polynomial in $L[y].$

Then, such a polynomial $P(x,y)=\sum_{k=0}^nP_k(x)y^k$ satisfies: $\forall k\quad P_k(a)=0$ by 2., hence $\forall k\quad P_k(x)=0$ by 1., i.e. $P(x,y)=0,$ q.e.d.