I'm working through Atiyah & MacDonald, and there's an exercise basically asking you to fill in a certain step in Artin's construction of an algebraic closure for a given field. The question is (this is not a quote from the book as I haven't got it here, but it's the same idea):
Let $k$ be a field. Let $\Sigma$ be the set of irreducible monic polynomials over $k$. Then form the polynomial ring $A=k[\{x_f:f\in\Sigma\}]$, where the $x_f$ are indeterminates indexed by the irreducible polynomials in $\Sigma$. Define an ideal $\mathfrak{a}$ to be the ideal in $A$ generated by polynomials of the form $f(x_f)$, where $f\in\Sigma$. Show that $\mathfrak{a}$ is a proper ideal.
(The question then goes on to describe how you can use this to construct an algebraic closure of $k$.)
I think I've found an answer to this. However, it's a lot shorter than other answers I've seen online, so I think there must be a problem with it. Here it is:
My answer
It is sufficient to show that we never have an equation of the form:
$$
F_1f_1(x_{f_1})+\dots+F_nf_n(x_{f_n})=1
$$
where $f_1,\dots,f_n\in\Sigma$ and $F_1,\dots,F_n\in A$. But we know that there is some field extension $K$ of $k$ in which the polynomials $f_i$ have roots $\alpha_i$. Working in $K$, we substitute in $\alpha_i$ for $x_{f_i}$ in the above expression, and obtain:
$$0=1$$
which is clearly impossible, since $k\subset K$, so $K$ is not the one-element field. It follows that $\mathfrak a$ is a proper ideal of $A$.
What is wrong with this answer?
Answer
There is nothing wrong with your proof.
In fact, this is exactly the same argument that Serge Lang gives in his Algebra
at page 232 (following Artin's argument).
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