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?

  • $\begingroup$ @user26857 OK. The only proof I could find online involved induction on $n$ and passing to quotient rings. I suppose that's implicit in this answer (the existence of the field $K$ is proved inductively and by using quotient rings), so maybe that way just seemed purer or more beautiful to the author. (sierra.nmsu.edu/morandi/notes/algebraicclosure.pdf) $\endgroup$ – John Gowers Oct 12 '14 at 23:53
  • $\begingroup$ @Donkey_2009: Do you have Lang's Algebra? If you look at p. 231, he proves that for $f \in k[X], deg f>0$, then there exists extension $E \supset k$ in which $f$ has a root. Then if you have a finite number of polynomials over $k$ it is an easy induction to show that there is extension $E$ in which all of them have a root. I don't think taking quotients is necessary anywhere. $\endgroup$ – Manos Oct 12 '14 at 23:59
  • $\begingroup$ @Manos The construction of an extension $E/k$ in which an irreducible polynomial $f\in k[x]$ has a root is normally done by taking $E=k[x]/(f(x))$, with the obvious inclusion of $k$. Then, if we let $\alpha = x + (f(x)) \in E$, $\alpha$ is a root of $f$. $\endgroup$ – John Gowers Oct 13 '14 at 17:00
  • $\begingroup$ Of course. What i meant is once we have shown that, then no further quotients are needed. $\endgroup$ – Manos Oct 13 '14 at 17:10
  • $\begingroup$ Yes, that's right. In the link I posted in my first comment, the author doesn't use any results about field extensions where certain polynomials have roots, and has to construct the quotient rings explicitly at the end. Purely formally, this gives you a simpler proof, but a less intuitive one. $\endgroup$ – John Gowers Oct 13 '14 at 17:12

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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