Describe all sets of reals that can be the image of the real line under a polynomial with real coefficients.
Find the image of the real plane under the polynomial \[ x^2+(xy-1)^2. \]
Describe all sets of reals that can be the image of the real plane under a polynomial in two variables with real coefficients.
Solution
1. A constant polynomial \(p\in\RR[t]\) has image a singleton.
If \(p(t)=c\) for every \(t\in\RR\), then \[ p(\RR)=\{c\}. \] Conversely, every singleton occurs in this way.
2. Every nonconstant odd-degree polynomial \(p\in\RR[t]\) has image \(\RR\).
Let the leading term of \(p\) be \(at^d\), where \(a\neq0\) and \(d\) is odd. Then the limits of \(p(t)\) as \(t\to+\infty\) and \(t\to-\infty\) have opposite signs and infinite magnitude. Hence, for every \(r\in\RR\), there are \(u<v\) with \[ p(u)<r<p(v) \] or with the two inequalities reversed. Since \(p\) is continuous, the intermediate value theorem gives some \(t\in[u,v]\) such that \(p(t)=r\).
3. Every nonconstant even-degree polynomial \(p\in\RR[t]\) has image a closed half-line.
Write the leading term as \(at^d\), where \(d\) is even. If \(a>0\), then \[ p(t)\longrightarrow+\infty \] as \(t\to\pm\infty\). Choose \(R>0\) so large that \[ \abs{t}\geq R \quad\Longrightarrow\quad p(t)>p(0). \] By the extreme value theorem, \(p\) attains a minimum \(m\) on \([-R,R]\), and the displayed inequality shows that this is a global minimum. Continuity and the limit \(p(t)\to+\infty\) then imply \[ p(\RR)=[m,\infty). \]
If \(a<0\), apply the same argument to \(-p\). Thus \(p\) has a global maximum \(M\) and \[ p(\RR)=(-\infty,M]. \]
4. The sets occurring in part (a) are exactly\[ \boxed{ \{c\},\qquad \RR,\qquad [c,\infty),\qquad (-\infty,c] } \]with \(c\in\RR\).
Steps <1>1–<1>3 show that no other image is possible. Conversely, the four displayed types are realized respectively by \[ c,\qquad t,\qquad t^2+c,\qquad c-t^2. \] This proves part (a).
5. For\[ F(x,y)\coloneqq x^2+(xy-1)^2, \]one has\[ F(x,y)>0 \]for every \((x,y)\in\RR^2\).
Both summands are nonnegative. If their sum were \(0\), then \(x=0\) and \(xy-1=0\) simultaneously. But \(x=0\) gives \(xy-1=-1\), a contradiction.
6. The image in part (b) is\[ \boxed{(0,\infty)}. \]
Step 5 shows that the image is contained in \((0,\infty)\). Conversely, let \(r>0\) and choose \[ x=\sqrt r, \qquad y=\frac1{\sqrt r}. \] Then \(xy=1\), so \[ F(x,y)=x^2+(xy-1)^2=r. \] Thus every positive real number occurs. This proves part (b).
7. If \(P\in\RR[x,y]\) is nonconstant, then \(P(\RR^2)\) is an interval.
The plane \(\RR^2\) is connected and \(P\) is continuous. Therefore its image under \(P\) is connected. The connected subsets of \(\RR\) are precisely the intervals.
8. If \(P\in\RR[x,y]\) is nonconstant, then \(P(\RR^2)\) is unbounded.
Choose \(u,v\in\RR^2\) with \(P(u)\neq P(v)\). Define \[ q(t)\coloneqq P\bigl(u+t(v-u)\bigr). \] Then \(q\in\RR[t]\) and \[ q(0)=P(u)\neq P(v)=q(1), \] so \(q\) is nonconstant. Every nonconstant one-variable polynomial is unbounded in absolute value as \(\abs{t}\to\infty\). Hence \(q(\RR)\) is unbounded, and since \[ q(\RR)\subseteq P(\RR^2), \] the latter set is unbounded as well.
9. Therefore the image of a nonconstant polynomial \(P\in\RR[x,y]\) is one of\[ \RR, \qquad [c,\infty), \qquad (c,\infty), \qquad (-\infty,c], \qquad (-\infty,c) \]for some \(c\in\RR\).
By step 7 the image is an interval, and by step 8 it is unbounded. An interval unbounded in both directions is \(\RR\). If it is unbounded only above, its finite infimum \(c\) is either attained or not attained, giving respectively \([c,\infty)\) or \((c,\infty)\). The case of an interval unbounded only below is analogous.
10. Every set listed in step 9 occurs, and constant polynomials add exactly the singleton images.
The polynomial \(x\) has image \(\RR\). The polynomials \[ x^2+c \qquad\text{and}\qquad c-x^2 \] have images \([c,\infty)\) and \((-\infty,c]\). By step 6, \[ c+F(x,y) \qquad\text{and}\qquad c-F(x,y) \] have images \((c,\infty)\) and \((-\infty,c)\). Finally, the constant polynomial \(c\) has image \(\{c\}\).
Thus the sets occurring in part (c) are exactly \[ \boxed{ \{c\},\quad \RR,\quad [c,\infty),\quad (c,\infty),\quad (-\infty,c],\quad (-\infty,c) } \] with \(c\in\RR\).
11. Q.E.D.
Step 4 proves part (a), step 6 proves part (b), and step 10 proves part (c).